How to Crack Chemistry by Chem Lovers

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Contents Chapter 1 .......................................................................... 13 Chemical Arithmatic ......................................................... 13 Laws of chemical combination ......................................... 39 Atomic, Molecular and Equivalent masses .......... 66 The mole concept ...................................................... 151 Percentage composition & Molecular formula .. 186 Chemical stoichiometry ............................................. 196 Chapter 2. ....................................................................... 277 Discovery and Properties of anode, cathode rays neutron and Nuclear structure ...................................................... 277 Chapter 3. ....................................................................... 810 Chemical Bonding ........................................................... 810 Electrovalent bonding ................................................... 810 Covalent bonding ........................................................ 886 Co-ordinate or Dative bonding ............................... 976 Dipole moment ............................................................ 991 Polarisation and Fajan's rule ................................ 1029 Overlaping-  and - bonds ................................. 1061 Hybridisation ............................................................... 1082 O .................................................................................... 1220 O .................................................................................... 1220 Resonance .................................................................. 1248

VSEPR Theory .......................................................... 1260 C..................................................................................... 1286 Xe................................................................................... 1286 F .................................................................................. 1286 F .................................................................................. 1286 Molecular orbital theory .......................................... 1313 Hydrogen bonding .................................................... 1413 Types of bonding and Forces in solid .............. 1468 Chapter 4. .................................................................. 1500 Solutions ...................................................................... 1500 Solubility ..................................................................... 1500 Method of expressing concentration of solution ....................................................................................... 1506 Colligative properties ............................................... 1674 Lowering of vapour pressure................................ 1686 Ideal and Non-ideal solution ................................. 1745 Azeotropic mixture .................................................... 1778 Osmosis and Osmotic pressure of the solution ....................................................................................... 1784 Elevation of boiling boint of the solvent .......... 1857 Depression of freezing point of the solvent .... 1886 Colligative properties of electrolyte..................... 1921

Abnormal molecular mass ..................................... 1955

Chapter 1 Chemical Arithmatic 1. One fermi is (a)

cm

(b)

cm

(c)

cm

(d)

cm

Ans.

(a)

cm

2. A picometre is written as (a)

m

(b)

m

(c)

m

(d)

m

Ans. (d)

m

3. One atmosphere is equal to (a) 101.325 K pa (b) 1013.25 K pa (c)

Nm

(d) None of these

Ans. (a) 101.325 K pa 4. Dimensions of pressure are same as that of (a) Energy (b) Force (c) Energy per unit volume (d) Force per unit volume Ans.

(c)

Pressure

Energy

5. The prefix (a) Giga (b) Nano (c) Mega (d) Exa Ans. (d) Exa

per

unit

is

volume

6. Given the numbers : 161cm, 0.161cm, 0.0161 cm. The number of significant figures for the three numbers are (a) 3, 4 and 5 respectively (b) 3, 3 and 3 respectively (c) 3, 3 and 4 respectively (d) 3, 4 and 4 respectively Ans. (b) 3, 3 and 3 respectively 7. Significant figures in 0.00051 are

(a) 5 (b) 3 (c) 2 (d) 4 Ans. (c) 2 8. Which of the following halogen can be purified by sublimation (a) (b)

(c) (d) Ans. (d) 9. Difference in density is the basis of (a) Ultrafiltration (b) Molecular sieving (c) Gravity Separation (d) Molecular attraction Ans. (c) Gravity Separation

10. Which of the following elements of matter would best convey that there is life on earth (a) Oxygen (b) Hydrogen (c) Carbon (d) Iron Ans. (c) Carbon

The compound which is added to table

11.

salt for maintaining proper health is (a) KCl (b) (c) (d) Ans. (c) 12. Which of the following contains only one element

(a) Marble (b) Diamond (c) Glass (d) Sand Ans. 13. In

(b) Diamond known

number is of (a) Metals (b) Non-metals

elements,

the

maximum

(c) Metalloids (d) None of these Ans. (a) Metals 14. Which one of the following is not an element (a) Diamond (b) Graphite (c) Silica (d) Ozone

Ans. (c) Silica 15. A mixture of separated by (a) Distillation (b) Crystallization (c)

Sublimation

(d) Adding aceitic acid Ans. (b) Crystallization

and

can be

16. A

mixture

of

methyl

alcohol

acetone can be separated by (a) Distillation (b) Fractional distillation (c) Steam distillation (d) Distillation under reduced pressure Ans. (b) Fractional distillation

and

17. In the final answer of the expression . The number of significant figures is (a) 1 (b) 2 (c) 3 (d) 4 Ans. (b)

Least precise terms i.e., 9.0 has only two significant figures.Hence, final answer will have two significant figures. 18.81.4 g sample of ethyl alcohol contains 0.002 g of water. The amount of pure ethyl

alcohol

to

the

significant figures is (a) 81.398 g (b) 71.40 g

proper

number

of

(c) 91.4 g (d) 81 g Ans.

(a)

Pure

ethyl .

19. The unit J (a) (b) (c) (d) None of these

is equivalent to

alcohol

Ans. (a) JPa–1; Unit of work is Joule and unit of pressure is Pascal. Dimension of

Joule i.e. work

So, JPa–1

.

20.

From

the following masses, the

one which is expressed nearest to the milligram is (a) 16 g (b) 16.4 g (c) 16.428 g (d) 16.4284 g Ans. (c) 16.428 g

21. The number of significant figures is (a) 23 (b) 3 (c) 4 (d) 26 Ans. (b) 3 22. (a)

The prefix zepto stands for

in

(b) (c) (d) Ans. (d) 1 zepto 23. (a) 2 (b) 5 (c) 6 (d) 4

The significant figures in 3400 are

Ans. (a) As we know that all non zero unit are significant number. Therefore significant figure is 2. 24.

The number of significant figures

in 6.0023 are (a) 5 (b) 4 (c) 3 (d) 1

Ans. (a) Number of significant figures in 6.0023 are 5 because all the zeroes stand between two non zero digit are counted towards significant figures. Given

25.

,

, Significant figures in and (a) 2, 2, 1 (b) 2, 3, 4

are respectively

,

(c) 4, 2, 1 (d) 4, 2, 3 18. Ans. (b) Given

& the

,

In decimal

initial zeros after

point

are

not

significant.

Therefore, significant figures in are 2. Similarly in are

3

as

significant. In

in

this

significant figures case

final

zero

is

all the zeroes are

significant hence, in

significant figures are

4. 26.

The number of significant figures

in 60.0001 is (a) 5 (b) 6 (c) 3 (d) 2

Ans. (b) All the zeroes between two non zero digit are significatn. Hence in 60.0001 significant figures is 6. 27.

A sample was weighted using two

different balances. The result’s were (i) 3.929 g (ii) 4.0 g. How would the weight of the sample be reported (a) 3.929 g (b) 3 g

(c) 3.9 g (d) 3.93 g Ans. (d) Round off the digit at 2nd position of decimal 3.929 = 3.93.

Laws of chemical combination 1. Which

of

the

following

pairs

of

substances illustrate the law of multiple proportions (a) CO and CO2 (b) (c) (d) Ans. (a) CO and CO2

2. 1.0 g of an oxide of A contained 0.5 g of

A.

4.0

g

of

another

oxide

of

A

contained 1.6 g of A. The data indicate the law of (a) Reciprocal proportions (b) Constant proportions (c) Conservation of energy (d) Multiple proportions Ans. (d) Multiple proportions

3. Among

the

following

pairs

of

compounds, the one that illustrates the law of multiple proportions is (a) (b) (c) (d) Ans. (c)

4. The percentage of copper and oxygen in samples of

obtained by different

methods were found to be the same. This illustrates the law of (a) Constant proportions (b) Conservation of mass (c) Multiple proportions (d) Reciprocal proportions Ans. (a) Constant proportions

5. Two

samples

separately

reduced

of

lead

to

oxide

metallic

were

lead

by

heating in a current of hydrogen. The weight of lead from one oxide was half the weight of lead obtained from the other oxide. The data illustrates (a) Law of reciprocal proportions (b) Law of constant proportions (c) Law of multiple proportions

(d) Law of equivalent proportions Ans. (c) Law of multiple proportions 6. Chemical

equation

is

according to the law of (a) Multiple proportion (b) Reciprocal proportion (c) Conservation of mass (d) Definite proportions Ans. (c) Conservation of mass

balanced

7. Avogadro number is

(a) Number of atoms in one gram of element (b) Number of millilitres which one mole of a gaseous substances occupies at NTP (c) Number of molecules present in one gram molecular mass of a substance (d) All of these

Ans. (c) Number of molecules present in one gram molecular mass of a substance 8. Different propartions of oxygen in the

various oxides of nitrogen prove the

(a) Equivalent proportion (b) Multiple proportion (c) Constant proportion (d) Conservation of matter

Ans. (b) Multiple proportion 9. Two elements X and Y have atomic

weights of 14 and 16. They form a series of compounds A, B, C, D and E in which the same amount of element X, Y is present in the ratio 1 : 2 : 3 : 4 : 5. If the compound A has 28 parts by weight of X and 16 parts by weight of Y, then the

compound of C will have 28 parts weight of X and (a) 32 parts by weight of Y (b) 48 parts by weight of Y (c) 64 parts by weight of Y (d) 80 parts by weight of Y Ans. (b) 48 parts by weight of Y 10.

Carbon and oxygen combine to

form two oxides, carbon monoxide and

carbon dioxide in which the ratio of the weights

of

carbon

and

oxygen

is

respectively 12 : 16 and 12 : 32. These figures illustrate the (a) Law of multiple proportions (b) Law of reciprocal proportions (c) Law of conservation of mass (d) Law of constant proportions Ans. (a) Law of multiple proportions

11. A sample of calcium carbonate has the following percentage composition :

Ca = 40%; C = 12%; O = 48%If the law of constant proportions is true, then the weight of calcium in 4 g of a sample of calcium carbonate obtained from another source will be (a) 0.016 g (b) 0.16 g

(c) 1.6 g (d) 16 g Ans. (c) 1.6 g 12. n g of substance X reacts with m g of substance Y to form p g of substance R and q g of substance S. This reaction can be represented

as,

.

The

relation which can be established in the

amounts of the reactants and the products will be (a) (b) (c) (d) Ans. (b)

⇌

low of conservation of mass.

by

13. Which

of

the

following

is

the

best

example of law of conservation of mass (a) 12 g of carbon combines with 32 g of oxygen to form 44 g of (b) When 12 g of carbon is heated in a vacuum there is no change in mass

(c) A sample of air increases in volume when heated at constant pressure but its mass remains unaltered (d) The weight of a piece of platinum is the same before and after heating in air Ans. (a) 12 g of carbon combines with 32

g of oxygen to form 44 g of 14. The

law

of

multiple

proportions

illustrated by the two compounds

is

(a) Sodium chloride and sodium bromide (b) Ordinary water and heavy water (c) Caustic soda and caustic potash (d) Sulphur dioxide and sulphur trioxide Ans.

(d)

Sulphur

dioxide

and

sulphur

trioxide 15. In compound A, 1.00 g nitrogen unites with 0.57 g oxygen. In compound B, 2.00

g nitrogen combines with 2.24 g oxygen. In

compound C, 3.00 g nitrogen combines with 5.11 g oxygen. These results obey the following law (a) Law of constant proportion (b) Law of multiple proportion (c) Law of reciprocal proportion (d) Dalton's law of partial pressure

Ans.

(b)

Law

of

multiple

proportionHydrogen combines with oxygen to form

in which

16 g

of oxygen combine with 2 g of hydrogen. Hydrogen also combines with carbon to form

in

which

2

g

of

hydrogen

combine with 6 g of carbon. If carbon and oxygen combine together then they will do show in the ratio of

(a) 6 : 16 or 12 : 32 (b) 6 : 18 (c) 1 : 2 (d) 12 : 24 Ans. (a) 6 : 16 or 12 : 32 16. 2 g of hydrogen combine with 16 g of oxygen to form water and with 6 g of carbon to form methane. In carbon dioxide

12 g of carbon are combined with 32 g of oxygen. These figures illustrate the law of (a) Multiple proportions (b) Constant proportions (c) Reciprocal proportions (d) Conservation of mass Ans. (c) Reciprocal proportions

17. An element forms two oxides containing respectively 53.33 and 36.36 percent of oxygen. These figures illustrate the law of (a) Conservation of mass (b) Constant proportions (c) Reciprocal proportions (d) Multiple proportions Ans. (d) Multiple proportions

18.After

a

chemical

reaction,

the

total

mass of reactants and products (a) Is always increased (b) Is always decreased (c) Is not changed (d) Is always less or more Ans. (c) Is not changed 19. A

sample

of

pure

carbon

dioxide,

irrespective of its source contains 27.27%

carbon

and

72.73%

oxygen.

The

support (a) Law of constant composition (b) Law of conservation of mass (c) Law of reciprocal proportions (d) Law of multiple proportions Ans. (a) Law of constant composition

data

20.

The law of definite proportions is

not applicable to nitrogen oxide because

(a)

Nitrogen

atomic

weight

is

constant (b) Nitrogen molecular weight is variable (c) Nitrogen equivalent weight is variable (d) Oxygen atomic weight is variable

not

Ans. (c) Nitrogen

equivalent

weight

is

variable 21. Which one of the following pairs of compounds illustrates the law of multiple proportion (a) (b) MgO, (c) (d)

Ans. (d)

Atomic, Molecular and Equivalent masses 1. Which property of an element is always a whole number (a) Atomic weight (b) Equivalent weight (c) Atomic number (d) Atomic volume Ans. (c) Atomic number

2. Which one of the following properties of an element is not variable (a)

Valency

(b) Atomic weight (c) Equivalent weight (d) All of these Ans. (b) Atomic weight 3. The

modern

based on

atomic

weight

scale

is

(a) (b) (c) (d) Ans. (a) 4. 1 amu is equal to (a) (b) (c)

(d)

kg

Ans. (a) 5. Sulphur

forms . The

sulphur in (a)

is

8 g/mole

(b) 16 g/mole (c) 64.8 g/mole (d) 32 g/mole

the

chlorides

equivalent mass

of

Ans. (b) The atomic weight of sulphur =32 In

valency of sulphur =2

So equivalent mass of sulphur

.

6. The sulphate of a metal M contains 9.87% of M. This sulphate is isomorphous . The atomic weight of M

with is (a) (b) 36.3

40.3

(c) 24.3 (d) 11.3 Ans.

(c)

As

the

given

sulphate

is

would

be

isomorphous with its

formula

.m is the atomic weight of M, molecular

weight

of

Hence % of M

(given)

or or

or

.

7. When 100 ml of 1

solution and

10 ml of 10

solution are mixed together, the

resulting solution will be

(a)Alkaline (b) Acidic (c) Strongly acidic (d) Neutral Ans. (d) For NaOH, For

Hence,

.

,

8. In chemical scale, the relative mass of the

isotopic

mixture

of

oxygen

atoms

is assumed to be equal to (a)

16.002

(b) 16.00 (c) 17.00 (d) 11.00 Ans. (b) 16.00

9. For

preparing

0.1

N

solution

of

a

compound from its impure sample of which the percentage purity is known, the weight of the substance required will be (a) More than the theoretical weight (b) Less than the theoretical weight (c) Same as the theoretical weight (d) None of these Ans. (a) More than the theoretical weight

10. 1 mol of (a)

contains atoms of H

(b) 4 g atom of Hydrogen (c)

molecules of

(d) 3.0 g of carbon Ans. (b) 1 mole of

contains 4 mole of

hydrogen atom i.e. 4g atom of hydrogen. 11. In

the

reaction ,

the

equivalent weight of

M) is equal to (a) M (b) (c) (d) Ans. (a)

(mol. wt. =

12. When

potassium

permanganate

is

titrated against ferrous ammonium sulphate, the

equivalent

weight

permanganate is (a)

Molecular weight /10

(b) Molecular weight /5 (c) Molecular weight /2 (d) Molecular weight Ans. (b)

of

potassium

13. Boron has two stable isotopes, (19%) and that

should

(81%). The atomic mass appear

periodic table is (a) 10.8 (b) 10.2 (c) 11.2 (d) 10.0

for

boron

in

the

Ans. (a) Atomic mass

14. What is the concentration of nitrate ions if equal volumes of 0.1 are mixed together (a)

0.1 M

(b) 0.2 M (c) 0.05 M

and 0.1

(d) 0.25 M Ans. (c) 0.1M

will react with 0.1M

NaCl to form volume

. But as the

doubled,

conc.

of

. 15. Total number of atoms represented by the compound CuSO4.5H2O is (a) (b) 21

27

(c) 5 (d) 8 Ans. (b) 21 16. 74.5 g of a metallic chloride contain 35.5 g of chlorine. The equivalent weight of the metal is (a) (b) 35.5 (c) 39.0

19.5

(d) 78.0 Ans. (c) wt. of metallic chloride wt. of chlorine = 35.5 wt. of metal Equivalent

weight

of

metal

17. 7.5 grams of a gas occupy 5.8 litres of volume at STP the gas is (a) (b) (c) (d) Ans. (a) 22.4L =

5.8L of gas has mass "

"

"

So molecular weight = 29 So, molecular formula of compound is NO 18.The number of atoms in 4.25 g of is approximately (a) (b) (c) (d)

Ans. (d)

17gm

contains

molecules of 4.25gm

contains =

No. of atoms

.

19. One litre of a gas at STP weight 1.16

g it can possible be (a) (b) (c)

(d) Ans. (a)

1L of gas at S.T.P. weight

1.16g 22.4 L of gas at S.T.P. weight

This molecular weight indicates that given compound is

.

20. The vapour density of a gas is 11.2. The volume occupied by 11.2 g of the gas at ATP will be (a)

11.2 L

(b) 22.4 L (c) 1 L (d) 44.8 L Ans. (a) Molecular weight

22.4gm of gas occupies 22.4L at S.T.P. 11.2gm of gas occupies . 21. Equivalent weight of crystalline oxalic acid is (a) (b) 63 (c) 53

30

(d) 45 Ans. (b) Equivalent weight

Molecular weight of

.

22. The equivalent weight of an element is 4. Its chloride has a V.D 59.25. Then the valency of the element is (a) (b) 3

4

(c) 2 (d) 1 Ans. (b) Valency of the element

=3. 23. 1.25

g of a solid dibasic acid is

completely neutralised by 25 ml of 0.25 molar the acid is

solution. Molecular mass of

(a) 100 (b) 150 (c) 120 (d) 200 Ans. (d) Molarity

Molecular weight

.

24. The oxide of a metal has 32% oxygen.

Its equivalent weight would be (a)

34

(b) 32 (c) 17 (d) 8 Ans. (c) Let weight of metal oxide = 100gm Weight of oxygen

= 32gm

weight of metal

Equivalent weight of oxide . 25.

The mass of a molecule of water

is

kg

(a) (b) (c)

kg kg

(d)

kg

Ans. (a)

molecules has mass

1 molecules has mass

. 26. (a) (b)

1.24 gm P is present in 2.2 gm

(c) (d) Ans. (a) Choice (a) is

P is present in 220gm 1.24gm

27.

The

elements

A

P

is

atomic and

B

present

in

weights

of

are

and

40

=

two 80

respectively. If x g of A

contains y atoms,

how many atoms are present in 2x g of B (a) (b) (c) y (d) 2y Ans. (c) Number of moles of

A

Number of atoms of

A

(say)

Or Number of moles of B Number

28.

of

of

Assuming fully decomposed, the

volume of

released at STP on heating

9.85g of

(Atomic mass of Ba=137)

will be (a)

atoms

0.84 L

B

(b) 2.24 L (c) 4.06 L (d) 1.12 L Ans. (d) Molecular weight of =197 197gm produces 22.4L at S.T.P. 9.85gm produces S.T.P.

at

29.

If

is

Avogadro’s

number

then

number of valence electrons in 4.2 g of nitride ions (a) 2.4 (b) 4.2 (c) (d)

Ans. (a) 14 gm

ions have

valence electrons4.2gm of

30.

The weight of is

(a)

41.59 g

(b) 415.9 g (c) 4.159 g (d) None of these

ions have

molecules of

Ans. (c) [ Molecular weight of ] molecules has weight molecules has weight

I. Rearrange the following (I to IV) in the order of increasing masses and choose the correct answer from (a), (b), (c) and (d) (Atomic mass: N=14, O=16, Cu=63).

1 molecule of oxygen II. 1 atom of nitrogen III.

g molecular weight of oxygen

IV.

g atomic weight of copper

(a) II HBr > HI Ans.(a)Bond strength decreases as the size of the halogen increases from F to I.

43. Which one has a pyramidal structure

(a)

CH 4

(b)

NH 3

(c)

H2O

(d)

CO2

Ans. (b) NH has pyramidal structure, yet 3

nitrogen is

sp 3

hybridised. This is due to

the presence of lone pair of electron.

44. Among the following the pair in which

the two species are not isostructural is (a) (b) (c) (d)

BH 4−

PF6−

SiF4

IO3−

and and

NH 4+

SF6

and

SF4

and

XeO3

Ans. (c) SiF has symmetrical tetrahedral 4

shape which is due to

sp 3

hybridization of

the central sulphur atom in its excited state

configuration.

SF4

has distorted tetrahedral

or Sea- Saw geometry which arise due to sp 3 d

hybridization of central sulphur atom and

due to the presence of lone pair of electron in one of the equatorial hybrid orbital. 45. The maximum number of 90° angles

between bond pair-bond pair of electrons is observed in

(a) (b) (c) (d)

dsp 2

hybridization

sp 3 d

dsp 3

hybridization

hybridization

sp 3 d 2

hybridization

Ans. (d)

O

O

dsp2 hybridization sp3d hybridization (Four 90° (Six 90° angle angles between between bond bond pair and pair and bond bond pair) pair)

O

sp3d2 hybridization (Twelve 90° angle between bond pair and bond pair)

Molecular orbital theory 1. Bond

order

is

a

concept

in

the

molecular orbital theory. It depends on the number of electrons in the bonding and antibonding orbitals. Which of the following statements is true about it ? The bond order (a) Can have a negative quantity (b) Has always an integral value

(c) Can assume any positive or integral or fractional value including zero (d) Is a non zero quantity Ans. (a)Can have a negative quantity 2. The bond order of

(a) 1 (b) 2 (c) 2.5

NO

molecule is

(d) 3 Ans.(c)

=

B.O. =

8 −3 5 = = 2 .5 2 2

No. of bonding e − − No. of antibondin g e − 2

.

3. When two atomic orbitals combine they

form (a) One molecular orbital (b) Two molecular orbital (c) Three molecular orbital

(d) Four molecular orbital Ans. (b) One bonding M.O. and one antibonding M.O. 4. Which of the following species is the

least stable (a) (b) (c)

O2

O 2−2

O 2+1

(d)

O 2−1

Ans. (b) O

2− 2

is least stable.

5. The bond order is maximum in

(a) (b) (c) (d)

O2

O 2−1

O 2+1

O 2−2

Ans. (c)B.O. of B.O. of

O 2+1

O2

is 2, B.O. of

is 2.5 and of

O 22 −

O 2−1

is 1.5,

is 1.

6. Which of the following compounds of

boron does not exist in the free form (a)

BCl3

(b)

BF3

(c)

BBr3

(d)

BH 3

Ans. (d)Hydride of boron does not exist in

BH 3

form. It is stable as its dimer di borane

(B2 H 6 )

.

7. Molecular orbital theory was developed

mainly by (a) Pauling (b) Pauling and Slater (c) Mulliken

(d) Thomson Ans. (c)Mulliken 8. The bond order of a molecule is given

by (a) The difference between the number of electrons

in

bonding

and

antibonding

orbitals (b) Total number of electrons in bonding and antibonding orbitals

(c) Twice number

the of

difference

electrons

in

between

the

bonding

and

antibonding electrons (d) Half the difference between the number of electrons in bonding and antibonding electrons Ans.

(b)Total

number

of

electrons

bonding and antibonding orbitals

in

9. Oxygen molecule is paramagnetic

because (a) Bonding electrons are more than antibonding electrons (b) Contains unpaired electrons (c) Bonding electrons are less than antibonding electrons (d) Bonding electrons are equal to antibonding electrons

Ans. (c)Bonding electrons are less than antibonding electrons 10. Which one is paramagnetic from the

following (a) (b)

O 2−

NO

(c) Both (a) and (b) (d)

CN

−

Ans. (c)

O 2− (2  8 + 1 = 17 )

has

odd

number

of

electrons and hence it is paramagnetic. All the

remaining

CN − (6 + 7 + 1 = 14 )

NO (7 + 8 = 15 )

molecules/ions,

diamagnetic

has odd number of electrons and

hence it is paramagnetic. 11. The bond order in

(a) 1 (b) 2

i.e.,

N 2+

ion is

(c) 2.5 (d) 3 Ans. (c) B.O. = No. of N

b

− No. of N a 5 = = 2 .5 2 2

.

12. Out of the following which has smallest

bond length (a) (b) (c)

O 2+

O 2−

O2

(d) Ans. (b)Bond order of

O 2+

is highest so its

bond length is smallest. 13. Which of the following molecule is

paramagnetic (a) Chlorine (b) Nitrogen (c) Oxygen

(d) Hydrogen Ans. (c) Oxygen is paramagnetic due to the presence of two unpaired electron :

O 2 =  (1s) 2   (1s) 2  (2 s) 2   (2 s) 2

 (2 p x )2  (2 p y )2  (2 p x )2   (2 p y )1   (2 p z )1

14. Which molecule has the highest bond

order (a)

N2

(b)

Li2

(c)

He 2

(d)

O2

Ans.

(a)

N2

15. The molecular electronic configuration of

H 2−

ion is

(a)

( 1s )2

(b)

( 1s )2 ( x 1s )2

(c)

( 1s )2 ( x 1s )

1

(d)

( 1s )3

Ans. (c) ( 1s) ( 1s ) 2

16. The

x

1

paramagnetic

nature

of

oxygen

molecule is best explained on the basis of

(a)

Valence bond theory

(b) Resonance (c) Molecular orbital theory

(d) Hybridization Ans. (c) Molecular orbital theory 17. In

which

case

the

minimum between carbon and nitrogen (a)

CH 3 NH 2

(b)

C6 H 5 CH = NOH

(c)

CH 3 CONH 2

bond

length

is

(d)

CH 3 CN

Ans. (d)In

CH 3 CN

bond order between C and

N is 3 so its bond length is minimum. 18. Which one of the following species is

diamagnetic in nature (a)

He 2+

(b) H2 (c)

H 2+

(d)

H 2−

 (1s)

Ans. (b)

*

 (1s) B.O. Magnetic nature

He 2+

H2

H 2+

H 2−









1

1 2

 1 2

D

P

P

 1 2 P

(P = Paramagnetic, D = Diamagnetic) 19. Which one of the following oxides is

expected exhibit paramagnetic behaviour (a)

CO2

(b)

SO 2

(c)

ClO2

(d)

SiO 2

Ans. (c)Due to unpaired

e − ClO2

is

paramagnetic. 20. The bond order in

(a) 1 (b) 2 (c) 3

N2

molecule is

(d) 4 Ans. (c)The Bond order in 3,

N  N



Here,

Nb = 2 + 4 + 2 = 8

and

N2

molecule is

Na = 2

B.O. = ( 8 − 2) / 2 = 3.

21. Which one is paramagnetic and has the

bond order 1/2 (a) O

2

(b)

N2

(c)

F2

(d)

H 2+

Ans. (d) H has the bond order + 2

1 2

, it has

only one electron so it will be paramagnetic. 22. When two atoms of chlorine combine to

form one molecule of chlorine gas, the energy of the molecule (a) Greater than that of separate atoms (b) Equal to that of separate atoms

(c) Lower than that of separate atoms (d) None of the above statement is correct Ans. (c)When bond forms between two atom then their energy get lower than that of separate atoms because bond formation is an exothermic process. 23. An atom of an element A has three

electrons in its

outermost shell and that of B has six electrons

in

the

outermost

shell.

The

formula of the compound between these two will be (a)

A3 B4

(b)

A 2 B3

(c)

A3 B 2

(d)

A2 B

Ans.

(b)Valency of A is 3 while that of

B is 2 so according to Criss Cross rule the formula of the compound between these two will be 24. The

benzene is

(b) Two

.

bond order of individual carbon-

carbon bonds in

(a) One

A 2 B3

(c) Between 1 and 2 (d) One and two alternately Ans. (c)Due to resonance bond order of

C −C

bonds in benzene is between 1 and 2.

25. PCl exists but 5

NCl 5

does not because

(a) Nitrogen has no vacant d-orbitals (b)

NCl 5

is unstable

(c) Nitrogen atom is much smaller

(d) Nitrogen is highly inert Ans. (a)Nitrogen does not have vacant ‘d’orbitals so it can’t have +5 oxidation state i.e. the reason

PCl5

exists but

NCl5

does not.

26. Paramagnetism is exhibited by

molecules (a) Not attracted into a magnetic field (b) Containing only paired electrons (c) Carrying a positive charge

(d) Containing unpaired electrons Ans. (d)Molecules having unpaired electrons show paramagnetism. 27. Which

one

paramagnetic (a)

H2O

(b)

NO 2

(c)

SO 2

of

the

following

is

(d)

CO2

Ans. (b) NO

2

has unpaired electrons so it

would be paramagnetic. 28. The

energy

of

a

2p

orbital

hydrogen atom is (a) Less than that of (b) More than that of (c) Equal to that of

2s

2s

2s

orbital orbital

orbital

except

(d) Double that of

2s

orbital

Ans. (b)More than that of

2s

orbital

29. In the electronic structure of acetic acid,

there are (a) 16 shared and 8 unshared electrons (b) 8 shared and 16 unshared electrons (c) 12 shared and 12 unshared electrons (d) 18 shared and 6 unshared electrons

Ans. (a)16 shared and 8 unshared electrons 30. Which of the following does not exist

on the basis of molecular orbital theory (a) (b) (c)

H 2+

He 2+

He 2

(d)

Li2

Ans. (c)Helium molecule does not exist as bond order of 31. In

P4 O10 ,

He 2 = 0

.

the number of oxygen atoms

attached to each phosphorus atom is (a) 2 (b) 3

(c) 4 (d) 2.5 Ans.

(c)Structure of

is

P4 O10

O XVII.

O O

O

XXI. P

P XVIII.

P XXIII.

XX. O XXII.

O

O XVI. O

XIX. O P

XV. O

Each phosphorus is attached to 4 oxygen atoms.

32. Of the following statements which one

is correct (a) Oxygen and nitric oxide molecules are both paramagnetic because both contain unpaired electrons (b) Oxygen and nitric oxide molecules are both diamagnetic because both contain no unpaired electrons

(c) Oxygen

is

paramagnetic

because

it

contains unpaired electrons, while nitric oxide is diamagnetic because it contains no unpaired electrons (d) Oxygen

is

diamagnetic

because

it

contains no unpaired electrons, while nitric oxide is paramagnetic because it contains an unpaired electron

Ans. (a)Oxygen and nitric oxide molecules are

both

paramagnetic

because

both

contain unpaired electrons 33. According

to

the

theory, the bond order in

C2

(a) 0 (b) 1

molecule is

molecular

orbital

(c) 2 (d) 3 Ans. (c)B.O. of carbon

=

Nb − Na 8−4 = =2 2 2

.

34. The molecular orbital configuration of a

diatomic molecule is   2 py2  1s 2  * 1s 2  2 s 2  * 2 s 2  2 p x2  2   2 pz

Its bond order is

(a) 3 (b) 2.5 (c) 2 (d) 1 Ans. (a)B.O.

=

Nb − Na 10 − 4 = =3 2 2

.

35. The difference in energy between the

molecular orbital

formed and the combining atomic orbitals is called (a) Bond energy (b) Activation energy (c) Stabilization energy (d) Destabilization energy Ans. (c)Stabilization energy

36. According

to molecular orbital theory,

the paramagnetism of

O2

molecule is due to presence of

(a) Unpaired electrons in the bonding



molecular orbital (b) Unpaired electrons in the antibonding 

molecular orbital

(c) Unpaired

electron

molecular orbital

in

the

bonding



(d) Unpaired electrons in the antibonding 

molecular orbital

Ans.

(d)Unpaired

antibonding



molecular orbital

37. The bond order in

(a) 2 (b) 2.5 (c) 1.5

electrons

O 2+

is

in

the

(d) 3 Ans. (b) B.O.

=

Nb − Na 8−3 5 = = = 2.5 2 2 2

.

38. Which of the following is paramagnetic

(a) (b) (c) (d)

O2

CN

−

CO

NO +

Ans.(a)Electronic configuration of

O2

is

O2 =  (1s)2   (1s)2  (2 s)2   (2 s)2  (2 p x )2  (2 p y )2

 (2 p z )2   (2 p y )1   (2 p z )1

The molecule has two unpaired electrons So, it is paramagnetic 39. If

Nx

is the number of bonding orbitals of

an atom and

Ny

is the number of antibonding orbitals, then the molecule/atom will be stable if (a)

Nx  Ny

(b)

Nx = Ny

(c)

Nx  Ny

(d)

Nx  Ny

Ans. (a) N

x

 Ny

40.Which of the following molecular orbitals

has two nodal planes (a)  2s (b)  2 p (c) (d)

y

 * 2py

 * 2p x

Ans. (c)  2 p has two nodal planes. *

y

41. The number of nodal planes

has (a) Zero (b) One (c) Two (d) Three Ans. (c)Two

' d'

orbital

42. Atomic number of an element is 26.

The element shows (a) Ferromagnetism (b) Diamagnetism (c) Paramagnetism (d) None of these Ans.(a)Element with atomic number 26 is

Fe. It is a ferromagnetic.

43. What is correct sequence of bond order

(a) (b) (c) (d)

O 2+  O 2−  O 2

O 2+  O 2  O 2−

O 2  O 2−  O 2+

O 2−  O 2+  O 2

Ans.(b) Correct Sequence of bond order is

O 2+  O 2  O 22 −

B.O – 2.5

2

1.5

44. Which bond is strongest

(a) (b) (c) (d)

F−F

Br − F

Cl − F

I−F

Ans. (a) Due to small bond length. 45. Which

of

paramagnetic

the

following

is

not

(a) (b) (c) (d)

S −2

N2

−

O 2−

NO

Ans. (a) S

−2

have all paired electrons so it is

diamagnetic. 46. Which one of the following molecules is

paramagnetic

(a)

CO2

(b)

SO 2

(c)

NO

(d)

H2O

Ans. (c)NO has 15 electrons. 47.

N 2−

N2

and

and

O 2−

O2

are converted into monoanions

respectively,

which

of

the

following

statements is wrong (a) In

N2,

the

(b) In

O2 ,

the

N − N

O−O

bond weakens bond order increases

(c) In O , bond length increases 2

(d)

N 2−

becomes diamagnetic

Ans. (b)In the conversion of order decreases.

O2

into

O 2−

bond

48. With increasing bond order, stability of

a bond (a) Remains unaltered (b)

Decreases

(c) Increases (d) None of these Ans. (c)Increases 49. Which is not paramagnetic

(a) (b) (c) (d)

O2

O 2+

O 22 −

O 2−

Ans. (c) O

2− 2

does not have any unpaired

electron so it is diamagnetic. 50. The

pairs in

number O 22 −

of

molecular

antibonding

electron

ion on the basis of molecular orbital theory is (a) 4 (b) 3 (c) 2 (d) 5 Ans.

(a) O

electron

2− 2

consist

pair

[1s

of and

four 2s

antibonding have

two

antibonding and

2 p x 2 py

have two antibonding

electron pair]. 51. The bond order of

(a) 1 (b) 2 (c)

1 2

(d)

1 4

He 2+

molecule ion is

Ans.

(c)The

electron’s

molecular orbitals is B.O.

=

52. Which

2 −1 1 = = 0 .5 2 2

(a) (b) (c)

ClO2

ClO 2−

NO 2

in

1s 2 , 2 s 1

.

one

paramagnetism

distribution

does

not

exhibit

(d)

NO

Ans. (b) ClO

− 2

has all paired electrons hence

it does not show paramagnetism. 53. In which of the following pairs the two

molecules have identical bond order (a) (b)

N 2 , O 22 +

N 2 , O 2−

(c)

N 2− , O 2

(d)

O 2+ ,

N2

Ans.(a)B.O. = 12 [ N

N2 =

1 6 [10 − 4 ] = = 3 2 2

;

b

− Na]

O 22 + =

1 6 [10 − 4 ] = = 3 . 2 2

54. The bond order is not three for

(a) (b) (c)

N 2+

O 22 +

N2

(d)

NO +

Ans. (a)B.O. for 55. In

H 2 O2

N 2+

(a)

90 o

(b)

101 o

(c)

103 o

b

− Na]

= 12 [9 − 4] = 52 = 2.5 .

molecule, the angle between the

two O – H planes is

= 12 [N

(d)

105 o

Ans. (a) H O 2

two

O−H

2

contain bond angle between

planes about

56. Which

(b) (c)

F−F

C–C

N – N

.

of the following molecule has

highest bond energy (a)

90 o

(d)

O–O

Ans. (c)Nitrogen molecule has highest bond energy due to presence of triple bond. 57. Which of the following species would be

expected paramagnetic (a) Copper crystals (b)

Cu +

(c) (d)

Cu + +

H2

Ans. (c) Cu

2+

= [ Ar18 ] 3 d 9 4 s 0

it has one unpaired

electron so it is paramagnetic. 58. Which of the following is correct for

triple bond (a) 3s (b) 1p, 2s

N2

(c) 2p, 1s (d) 3p Ans. (c)2p, 1s 59. In

which

of

the

molecules have bond order three and are isoelectronics (a) (b)

CN

−

, CO

NO + , CO

+

following

pairs

(c)

CN

−

,

(d) CO,

O 2+

O 2+

Ans.(a) CN

−

= 14

electrons ; CO

=14 electrons

B.O. = 12 [10 − 4] = 62 = 3 . 60.Which of the following is paramagnetic

(a) (b)

O 2+

CN

−

(c) CO (d)

N2

Ans. (a)B.O. = 12 [10 − 5] = 52 = 2.5 , paramagnetic 61. How many bonding electron pairs are

there in white phosphorous (a) 6 (b) 12

(c) 4 (d) 8 Ans. (a)

P P

P P

62. The atomicity of phosphorus is X and

the

PPˆ P

bond angle in

the molecule is Y. What are X and Y (a) X = 4, Y =

90 o

(b) X = 4, Y =

60 o

(c) X = 3, Y =

120 o

(d) X = 2, Y =

180 o

Ans. (b)X = 4, Y = 63. From

60 o

elementary

molecular

orbital

theory we can give the electronic

configuration

of

positive nitrogen molecular ion (a)

 (1s)2  (1s)2  (2 s)2  (2 s)2  (2 p)4  (2 p)1

the N 2+

as

singly

(b)

 (1s)2  (1s)2  (2 s)2  (2 s)2  (2 p)1 (2 p)3

(c)

 (1s)2  (1s)2  (2 s)2  (2 p)2  (2 p)4

(d)

 (1s)2  (1s)2  (2 s)2  (2 s)2  (2 p)2  (2 p)2

Ans. (a)  (1s)  (1s)  (2s)  (2s)  (2 p)  (2 p) 2

64. The



2

2



2

4

paramagnetic

1

property

of

the

oxygen molecule due to the presence of unpaired electorns present in

(a) (b)

( 2 p x )1

( 2 p x )1

(c)

( * 2 p y )1

(d)

( * 2 p y )1

(e)

( * 2 p z )1

Ans.

and

( * 2 p x )1

( 2 p y )1

and and

( * 2 p z )1

( 2 p y )1

and

(c)

oxygen which

and

( 2 p z )1

The came

can

orbital theory.

be

paramagnetic through

property

unpaired

explained

by

in

electron molecular

2px

Antibond ing

*

2pz* 2py* P P Pz x y

P P Pz x y

2Px bonding

So 2 unpaired of electron present in  2 p y* and  2 p *z

65. In

.

PO 43 −

ion, the formal charge on each

oxygen atom and

P −O

bond order respectively are

(a)

−0.75 , 1.25

(b)

−0.75 , 1.0

(c)

−0.75 , 0.6

(d)

−3, 1.25

number of bonds between atoms Ans. (a) Bond order = Total Total number of resonating structure

=

66. The bond order in

(a) Zero

CO 32 −

5 = 1 .25 4

ion between

C −O

is

(b) 0.88 (c) 1.33 (d) 2 Ans.(c)We know that carbonate ion has following resonating structures –

O

–

O C=O 

–

–

O

Bond order

O

=

O

C–O 

C – O–

–

O

Total number of bonds between atoms Total number of resonating structure

=

1 +1 + 2 4 = = 1 .33 3 3

.

67. The bond order of

(a) (b) (c) (d)

O 2+

N 2+

CN −

CO

NO +

Ans.(a) O

+ − 2 (15 e )

= K : K * ( 2 s) 2 ( * 2 s) 2 ( 2 p x ) 2

( 2 p y )2 ( 2 p z )2 ( * 2 p y )1 ( * 2 p z )0

is the same as in

Hence, bond order

=

1 (10 − 5) = 2 .5 2

N 2+ (13 e − ) = KK * ( 2 s) 2 ( * 2 s) 2 ( 2 p x ) 2

( 2 p y )2 ( 2 p z )1

Hence, bond order 68. Bond order of

(a) 2 (b) 1.5 (c) 3

O2

=

1 (9 − 4 ) = 2 .5 2

is

.

(d) 3.5 Ans. (a)Electronic configuration of

O2

is

O 2 = ( 1 s) 2 ( * 1 s) 2 ( 2 s) 2 ( * 2 s) 2 ( * 2 s) 2 ( 2 p z ) 2

( 2 p x2   2 p y2 ) ( * 2 p 1x   * 2 p 1y )

Hence bond order

=

1 N b − N a  2

=

1 [10 − 6] = 2 2

.

69. The total number of electron that takes

part in forming bonds in

N2

is

(a) 2 (b) 4 (c) 6 (d) 10 Ans. (c)Nitrogen form triple bond

N  N

In which 6 electron take part. 70. The bond length the species

are in the order

O 2 , O 2+

and

O 2−

of (a) (b) (c) (d)

O 2+  O 2  O 2−

O 2+  O 2−  O 2

O 2  O 2+  O 2−

O 2−  O 2  O 2+

Ans.(a) As bond order increase bond length decrease the bond order of species are

=

number of bonding electron - Number of a.b. electron 2

For

O 2+ =

O 2− =

O2 =

10 − 6 =2 2

;

10 − 5 = 2.5 2

10 − 7 = 1.5 2

So, bond order are

O 2+  O 2  O 2−

O 2+  O 2  O 2−

and bond length

.

71. According

which of the

to

molecular

orbital theory

following

statement

character regarding

and

about

bond

the

order

magnetic is

O 2+

(a) Paramagnetic and bond order< O (b) Paramagnetic and bond order> O (c) Dimagnetic and bond order< O (d) Dimagnetic and bond order> O Ans. (b)

correct

O2

:

2

2

 2py 2  * 2py1  1s 2 ,  * 1s 2 ,  2 s 2 ,  * 2 s 2 ,  2 p x 2  2pz 2  * 2pz1

2

2

Bond order (Two

=

10 − 6 = 2.0 2

unpaired

electrons

in

antibonding

molecular orbital) O

=

+ 2

2 * 1 2  2 py  2 py :  1s 2 ,  * 1s 2 ,  2 s 2 ,  * 2 s 2 ,  2 p x  2 * 0  2 pz  2 pz

Bond

order

10 − 5 = 2.5 2

(One

unpaired

electron

in

antibonding

molecular orbital so it is paramagnetic) 72. The bond order in NO is 2.5 while that

in

NO +

is 3. Which

of

the

following

statements

is

true

for

these two species (a) Bond length in

(b) Bond length in (c) Bond length in

NO +

NO

is equal to that in

is greater than in

NO +

is greater than in

(d) Bond length is unpredictable

NO +

NO

NO

Ans. (b)Higher the bond order, shorter will be the bond length, thus

NO +

having the

higher bond order that is 3 as compared to

NO

having bond order 2 so

NO +

has

shorter bond length. 73. Which of the following is diamagnetic

(a) Oxygen molecule (b) Boron molecule (c)

N 2+

(d) None Ans.

(d)Oxygen

molecule unpaired

(B 2 )

and

molecule N 2+

electron,

boron

(O 2 )

ion, all of them have hence

they

all

are

paramagnetic. 74. Bond energies in

as (a)

NO −  NO  NO +

(b)

NO  NO −  NO +

NO , NO +

and

NO −

are such

(c)

NO +  NO  NO −

(d)

NO +  NO −  NO

Ans. (c)Bond order of

3, 2.5

NO + , NO

and

and 2 respectively, bond energy

NO −

are



bond

order 75. Which of the following is paramagnetic

(a)

B2

(b)

C2

(c)

N2

(d)

F2

Ans.

(a)Paramagnetic

through unpaired electron. the

unpaired

electron

paramagnetism.

B2 →  1s 2 * 1s 2 ,  2 s 2 * 2 s 2 ,  2 p x 1 =  2 p y 1 (2 unpaired electron)

C2 →

 1s 2 * 1s 2 ,  2 s 2 * 2 s 2 ,  2 p x 2 . 2 p y 2 (No unpaired electron)

property

B2

arise

molecule have so

it

show

N2 →

 1s 2 * 1s 2 ,  2 s 2 * 2 s 2 ,  2 p x 2 ,  2 p y 2 2 p z 2 (No unpaired electron)

F2 → s 2 ,  * 1s 2 ,  2 s 2 ,  * 2 s 2 ,  2 p x 2 ,  2 p y 2 ,  2 p z 2 , (No unpaired electron)

 * 2 py 2 , * 2 pz 2

So only

B2

exist unpaired electron and

show the paramagnetism. 76. The paramagnetic molecule at ground

state among the following is

(a)

H2

(b)

O2

(c)

N2

(d)

CO

Ans. (b)

O2 →  1s ,  1s ,  2 s ,  2 s ,  2 p x 2

*

2

2

*

2

 2 py 2  * 2 py1 2

 2 pz 2  * 2 pz 2

So two unpaired electron found in ground

stage

paramagnetism.

by

which

it

O2

at

shows

77. Which has the highest bond energy

(a)

F2

(b)

Cl 2

(c)

Br2

(d)

I2

Ans. (b)Due to greater electron affinity has the highest bond energy.

Cl2

78. In

O 2− ,

O2

and

O 2−2

molecular species, the

total number of antibonding electrons respectively are (a) 7, 6, 8 (b) 1, 0, 2 (c) 6, 6, 6 (d) 8, 6, 8

Ans.(a)Molecular

orbital

electronic

configuration of these species are :

O2− (17 e − ) =

 1s 2 * 1s 2 ,  2 s 2 * 2 s 2 ,  2 p x 2 ,  2 p y 2 ,

 2 p z 2 ,  * 2 p y 2 * 2 p z 1

2

2

2 * 2 2 * 2 O2 (16 e) =  1s  1s ,  2 s  2 s ,  2 p x ,  2 p y ,

 2 p z 2 * 2 p y 1 * 2 p z 1

O 22 − (18 e ) =

 1s 2 * 1s 2 ,  2 s 2 * 2 s 2 ,  2 p x 2 ,  2 p y 2 ,

 2 p z 2 * 2 p y 2 * 2 p z 2

Hence

number

of

antibonding

electrons are 7,6,and 8 respectively. 79. Which

of

paramagnetic (a) (b) (c)

O2

O 22 +

O 22 −

the

following

is

not

(d)

O 2−

Ans. (c)Species with unpaired electrons is paramagnetic O 2−

O2

has 2 unpaired electrons,

has one unpaired,

electrons,

O 22 +

has zero unpaired

has one unpaired.

Which

80.

O 22 −

of

the

following

have maximum number of unpaired electrons (a)

O2

species

(b) (c) (d)

O 2+

O 2−

O 22 −

Ans. (a) O has 2 unpaired electron while 2

and while

O 2+

has one each unpaired electrons

O 2−

O 22 +

does

not

have

any

unpaired

electron. 81. The correct order in which the O – O

bond length

increases in the following is (a)

H 2 O2  O2  O3

(b)

O2  H 2 O2  O3

(c)

O2  O3  H 2 O2

(d)

O3  H 2 O2  O2

Ans.(c)

H −O −O−H

O O

O O

O

O

,O O = O,

O =O

Due to resonance in be in b/w

O =O

and

O −O

O3 O − O

bond length will

.

82. Correct order of bond length is

(a) (b) (c)

CO 32 −  CO 2  CO

CO 2  CO  CO 32 −

CO  CO 2  CO 32 −

(d) None of these

Ans. (a)From valency bond theory, bond order in CO, i.e. 2 while that of

−

+

:C O:

CO 32 −

is 3, that of

O=C =O

is

ion is 1.33. Since the

bond length increases as the bond order decreases, i.e.

CO  CO 2  CO 32 −

.

83. Which of the following is paramagnetic

(a)

N2

(b)

C2

(c) (d)

N 2+

O 22 −

Ans. (c) N

2

: KK (2 s)2  * (2 s)2  (2 p x )2  (2 py )2  (2 p z )2

(diamagnetic)

C2 : KK (2 s)2  * (2 s)2  (2 p x )2  (2 py )2

(diamagnetic)

N 2+ : KK (2 s)2  * (2 s)2  (2 p x )2  (2 py )2  (2 p z )2

(paramagnetic)

O22 − : KK (2 s)2  * (2 s)2  (2 p z )2  (2 p x )2  (2 py )2

 * (2 p x )2  * (2 py )2

84. Among

the following molecules which

one have smallest bond angle

(a)

NH 3

(b)

PH3

(diamagnetic)

(c)

H 2O

(d)

H 2 Sc

(e)

H2S

Ans. (d) NH

3

= 107 , PH3 = 93 , H 2O = 104 .5

H 2 Se = 91 , H 2 S = 92 .5

Hydrogen bonding

1. In

the following which bond will be

responsible for

maximun value of hydrogen bond (a)

O−H

(b)

N −H

(c)

S −H

(d)

F−H

Ans. (d)Hydrogen bonding will be maximum in

F-H

bond

due

electronegativity difference.

to

greater

2. In which of the following hydrogen bond

is present (a)

H2

(b) Ice (c) Sulphur (d) Hydrocarbon Ans. (b)Ice has hydrogen bonding.

3. In the following which has highest

boiling point (a)

HI

(b)

HF

(c)

HBr

(d)

HCl

Ans.(b)H – F has highest boiling point because it has hydrogen bonding.

4. Which contains hydrogen bond

(a)

HF

(b)

HCl

(c)

HBr

(d)

HI

Ans. (a) HF 5. Contrary

to

other

hydrogen fluoride is a

hydrogen

halides,

liquid because (a) Size of

F

atom is small

(b)

HF

is a weak acid

(c)

HF

molecule are hydrogen bonded

(d) Fluorine is highly reactive Ans. (c) HF molecule are hydrogen bonded 6. In the following which species does not

contain

sp 3

hybridization (a)

NH 3

(b)

CH4

(c)

H 2O

(d)

CO2

Ans. (d) CO is sp-hybridised 2

7. As a result of

sp

hybridization, we get

(a) Two mutual perpendicular orbitals

(b) Two orbitals at

180 o

(c) Four orbitals in tetrahedral directions (d) Three orbitals in the same plane Ans. (b) sp-hybridization gives two orbitals at

180 o

with Linear structure.

8. The reason for exceptionally high boiling

point of water is (a) Its high specific heat

(b) Its high dielectric constant (c) Low ionization of water molecules (d) Hydrogen bonding in the molecules of water Ans.

(d)Hydrogen bonding increases the

boiling point of compound. 9. Which

concept best explains that o-

nitrophenol is more

volatile than p-nitrophenol (a)Resonance (b) Hyperconjugation (c) Hydrogen bonding (d) Steric hindrence Ans. (c)o-Nitrophenol has intramolecular hydrogen bonding but p-Nitrophenol has intermolecular hydrogen bonding so boiling

point of p-Nitrophenol is more than oNitrophenol. 10. Which contains strongest

(a)

O − H ..... S

(b)

S − H ..... O

(c)

F − H ..... F

(d)

F − H ..... O

H−

bond

Ans.(c) The strongest hydrogen bond is in hydrogen fluoride because the power of hydrogen bond  electronegativity of atom and electronegativity



1 atomic size

So fluorine has maximum electronegativity and minimum atomic size. 11. Which of the following compound can

form hydrogen

bonds (a) (b)

CH 4

NaCl

(c)

CHCl3

(d)

H2O

Ans. (d) H O can form hydrogen bonds rest 2

CH 4

and

CHCl3

are organic compound having

no oxygen while NaCl has itself intraionic attraction in the molecule. 12. Of the following hydrides which has the

lowest boiling point (a)

NH 3

(b)

PH 3

(c)

SbH 3

(d)

AsH 3

Ans. (b) PH has the lowest boiling point 3

because it does not form Hydrogen bond. 13. The pairs of bases in

together by (a) Hydrogen bonds (b) Ionic bonds (c) Phosphate groups

DNA

are held

(d) Deoxyribose groups Ans. (a)Hydrogen bonds 14. Water has high heat of vaporisation due

to (a) Covalent bonding (b)

H −

bonding

(c) Ionic bonding (d) None of the above

Ans. (b)Hydrogen bonding increases heat of vaporisation. 15. In

which of the following compounds

does hydrogen bonding occur (a) (b) (c)

SiH 4

LiH

HI

(d)

NH 3

Ans. (d)Only

NH 3

forms H-bonds.

16. Which among the following compounds

does not show hydrogen bonding (a) Chloroform (b) Ethyl alcohol (c) Acetic acid

(d) Ethyl ether Ans. (d)Ethyl ether 17. Acetic acid exists as dimer in benzene

due to (a) Condensation reaction (b) Hydrogen bonding (c) Presence of carboxyl group (d) Presence of hydrogen atom at

−

carbon

Ans. (b)Hydrogen bonding 18. Which one among the following does

not have the hydrogen bond (a) Phenol (b) Liquid

NH 3

(c) Water (d) Liquid

HCl

Ans. (d)Liquid

HCl

19. The bond that determines the secondary

structure of a protein Is (a) Coordinate bond (b) Covalent bond (c) Hydrogen bond (d) Ionic bond

Ans. (c)Hydrogen bond 20. HCl is a gas but

HF

is a low boiling liquid.

This is because (a)

H−F

bond is strong

(b)

H−F

bond is weak

(c) Molecules

aggregate

hydrogen bonding (d)

HF

is a weak acid

because

of

Ans. (c)Molecules aggregate because of hydrogen bonding 21. The relatively high boiling point of

due to (a) Hydrogen bonding (b) Covalent bonding (c) Unshared electron pair on (d) Being a halogen acid

F

HF

is

Ans. (a)Hydrogen bonding 22. Water is liquid due to

(a) Hydrogen bonding (b) Covalent bond (c) Ionic bond (d) Vander Waals forces

F

Ans.

(a)Water

molecule

has

hydrogen

bonding so molecules get dissociated so it is liquid. 23. The

maximum

possible

number

hydrogen bonds in which an (a) 1 (b) 2

H2O

molecule can participate is

of

(c) 3 (d) 4 Ans.

(d)In

case

of

water,

five

water

molecules are attached together through four hydrogen bonding. 24. Hydrogen bonding is maximum in

(a) Ethanol (b) Diethyl ether

(c) Ethyl chloride (d) Triethyl amine Ans. (a)Ethanol 25. The hydrogen bond is strongest in

(a) Water (b) Ammonia (c) Hydrogen fluoride (d) Acetic acid

Ans.

(c)Hydrogen

bond

is

strongest

in

hydrogen fluoride. 26. The high boiling point of ethanol

(78 .2 o C)

compared to dimethyl ether (−23 .6 C) , though both having the o

same molecular formulae (a) Hydrogen bonding (b) Ionic bonding

C6 H 6 O

, is due to

(c) Coordinate covalent bonding (d) Resonance Ans. (a)Hydrogen bonding 27. Methanol and ethanol are miscible in

water due to (a) Covalent character (b) Hydrogen bonding character (c) Oxygen bonding character

(d) None of these Ans. (b)Hydrogen bonding character 28. B.P. of

H 2 O (100 o C )

and

H 2 S (−42 o C )

(a) Vander Waal's forces (b) Covalent bond (c) Hydrogen bond (d) Ionic bond

explained by

Ans. (c)Boiling point of that of

H2S

because

bonding while

H2S

29. Strength

of

H 2O

H 2O

is more than

forms hydrogen

does not. hydrogen

intermediate between (a) Vander Waal and covalent (b) Ionic and covalent (c) Ionic and metallic

bond

is

(d) Metallic and covalent Ans. (a)Vander Waal and covalent 30. In

which of the following compounds

intramolecular hydrogen bond is present (a) Ethyl alcohol (b) Water (c) Salicylaldehyde

(d) Hydrogen sulphide O C

H

O

+ −

H

Ans. (c)

Interamolecular

H-bonding.

31. Hydrogen

bonding

is

compounds containing hydrogen and (a) Highly electronegative atoms

formed

in

(b) Highly electropositive atoms (c) Metal atoms with d-orbitals occupied (d) Metalloids Ans. (a)Hydrogen bond is formed when hydrogen is attached with the atom which is highly electronegative and having small radius. 32. Which of the following compounds in

liquid state does not

have hydrogen bonding (a) (b)

H2O

HF

(c)

NH 3

(d)

C6 H 6

Ans. (a) H O 2

33. Compounds showing hydrogen bonding

among

HF, NH 3 , H 2 S

and

PH 3

are

(a) Only (b) Only (c) Only

HF, NH 3

HF

and

and

NH 3 , H 2 S

PH 3

NH 3

and

PH 3

(d) All the four Ans. (d)All the four 34. The high density of water compared to

ice is due to

(a) Hydrogen bonding interactions (b) Dipole-dipole interactions (c) Dipole-induced dipole interactions (d) Induced

dipole-induced

dipole

interactions Ans. (a)Water is dense than ice because of

hydrogen

structure of ice.

bonding

interaction

and

35. Ethanol and dimethyl ether form a pair

of functional isomers. The boiling point of ethanol is higher than that of dimethyl ether due to the presence of (a) Hydrogen bonding in ethanol (b) Hydrogen bonding in dimethyl ether (c)

CH 3

group in ethanol

(d)

group in dimethyl ether

Ans.

(a)Ethanol have hydrogen bonding

so its boiling point is higher than its isomer dimethyl ether. 36. Which of the following hydrogen bonds

are strongest in vapour phase (a)

HF − − − HF

(b)

HF − − − HCl

(c)

HCl − − − HCl

(d)

HF − − − HI

Ans.

(a)A

compound

electronegative

element

having will

maximum

form

strong

Hydrogen bond. 37. Which of the following shows hydrogen

bonding

(a)

NH 3

(b) P (c) As (d) Sb Ans. (a)Due to electronegativity difference of

N2

and

38. The

H2

,

NH 3

form hydrogen bond.

boiling point of a compound is

raised by

(a) Intramolecular hydrogen bonding (b) Intermolecular hydrogen bonding (c) Covalent bonding (d) Ionic covalent Ans.

(b)Intermolecular

hydrogen

bonding

compound contain more b.p. compare to intramolecular hydrogen bonding compound.

39. The

boiling

point

of

water

is

exceptionally high because (a) Water molecule is linear (b) Water molecule is not linear (c) There is covalent bond between H and

O (d) Water

molecules

hydrogen bonding

associate

due

to

Ans. (d)Water molecule contain hydrogen bonding. 40. NH 3 has a much higher boiling point than

PH 3

because

(a)

NH 3

has a larger molecular weight

(b)

NH 3

undergoes umbrella inversion

(c)

NH 3

forms hydrogen bond

(d)

NH 3

contains ionic bonds whereas

PH 3

contains covalent bonds Ans. (c)It contain intermolecular hydrogen bonding. 41. Which one has the highest boiling point

(a) Acetone (b)

Ethyl alcohol

(c) Diethyl ether (d) Chloroform Ans. (b)Ethyl alcohol has a intermolecular hydrogen bond. 42. Which of the following compounds has

the highest boiling point (a)

HCl

HBr

(b) (c)

H 2 SO 4

(d)

HNO3

Ans. (b) HBr 43. Which

of the following has minimum

melting point (a) CsF (b) HCl

(c) HF (d) LiF Ans. (b)HCl contain weak covalent bond. 44. Hydrogen bond energy is equal to

(a) 3 – 7 cals (b) 30 – 70 cals (c) 3 – 10 kcals

(d) 30 – 70 kcals Ans. (c)3 – 10 kcals 45. H 2 O is a liquid while

H2S

is gas due to

(a) Covalent bonding (b) Molecular attraction (c) H – bonding (d) H – bonding and molecular attraction

Ans.

(c)Due

to

intermolecular

hydrogen

bonding water molecules come close to each other and exist in liquid state. 46. H – bonding is maximum in

(a)

C6 H 5 OH

(b)

C6 H 5 COOH

(c)

CH 3 CH 2 OH

(d)

CH 3 COCH 3

Ans. (b)Due to greater resonance stabilization. 47. Select the compound from the following

which dissolves in water (a)

CCl4

(b)

CS 2

(c)

CHCl3

(d)

C2 H5 OH

Ans. (d) C H OH will dissolve in water because 2

5

it forms hydrogen bond with water molecule. 48. When two ice cubes are pressed over

each other, they unit to form one cube. Which of the following force

is

together

responsible

for

holding

them

(a) Vander Waal's forces (b) Hydrogen bond formation (c) Covalent attraction (d) Dipole–dipole attraction Ans. (b)In ice cube all molecules are held by inter molecular hydrogen bond. 49. Which

is

the

weakest

following types of bond

among

the

(a) Ionic bond (b) Metallic bond (c) Covalent bond (d) Hydrogen bond Ans. (d)Hydrogen bonding is developed due to inter atomic attraction so it is the weakest. 50. H-bond is not present in

(a) Water (b) Glycerol (c) Hydrogen fluoride (d) Hydrogen Sulphide Ans. (d)Hydrogen Sulphide

Types of bonding and Forces in solid

1. In a crystal cations and anions are held

together by (a)

Electrons

(b) Electrostatic forces (c) Nuclear forces

(d) Covalent bonds Ans. (b)In electrovalent crystal has cation and anion are attached by electrostatic forces. 2. In the following metals which one has

lowest probable interatomic forces (a)

Copper

(b) Silver (c) Zinc (d) Mercury Ans. (d)Mercury has very weak interatomic forces so it remains in liquid state. 3. In solid argon, the atoms are held

together by (a) Ionic bonds

(b) Hydrogen bonds (c) Vander Waals forces (d) Hydrophobic forces Ans.

(c)The melting and boiling points

of argon is low hence, in solid argon atoms are held together by weak Vander Waal’s forces. 4. Which one is the highest melting halide

(a) (b) (c) (d)

NaCl

NaBr

NaF

NaI

Ans. (c)NaF is the strongest ionic crystal so its melting point would be highest. 5. The

enhanced

metals is due to

force

of

cohesion

in

(a) The covalent linkages between atoms (b) The

electrovalent

linkages

between

atoms (c) The

lack

of

exchange

of

valency

energy

of

mobile

electrons (d) The

exchange

electrons Ans. (d)The exchange energy of mobile electrons

6. Which one of the following substances

consists of small discrete molecules (a)

NaCl

(b) Graphite (c) Copper (d) Dry ice Ans. (d)Dry ice

7. Which of the following does not apply

to metallic bond (a) Overlapping valency orbitals (b) Mobile valency electrons (c) Delocalized electrons (d) Highly directed bonds Ans. (d)Highly directed bonds 8. In melting lattice, structure of solid

(a) Remains unchanged (b) Changes (c) Becomes compact (d) None of the above Ans.

(b)Changes

9. Which of the following has the highest

melting point (a)

Pb

(b) Diamond (c) (d)

Fe

Na

Ans. (b)Diamond is the hardest substance it’s melting point would be highest. 10. In the formation of a molecule by an

atom (a) Attractive forces operate

(b) Repulsive forces operate (c) Both

attractive

and

repulsive

forces

operate (d) None of these Ans.(c)Bond is formed by attractive and repulsive forces of both the atoms. 11. Which has weakest bond

(a) Diamond

(b) Neon (Solid) (c)

KCl

(d) Ice Ans. (d)Ice 12. Which

of

the

following

weakest intermolecular forces (a)

He

exhibits

the

(b)

HCl

(c)

NH 3

(d)

H2O

Ans.

(a)Generally zero group elements

are linked by the

Vander Waal’s force.

Hence these show weakest intermolecular forces. 13. Glycerol

has

strong

bonding therefore it is

intermolecular

(a) Sweet (b) Reactive (c) Explosive (d) Viscous Ans. (d)Glycerol has a three OH group hence it is viscous in nature. 14. Among the following the weakest one is

(a) Metallic bond

(b) Ionic bond (c) Van der Waal's force (d) Covalent bond Ans.(c) Vander waal's forces is the weakest force of attraction. 15. Lattice energy of alkali metal chlorides

follows the order (a)

LiCl  NaCl  KCl  RbCl  CsCl

(b)

CsCl  NaCl  KCl  RbCl  LiCl

(c)

LiCl  CsCl  NaCl  KCl  RbCl

(d)

NaCl  LiCl  KCl  RbCl  CsCl

Ans. (a) LiCl  NaCl  KCl  RbCl  CsCl 16. In the following which molecule or ion

possesses electrovalent,

covalent

bond at the same time

and

co-ordinate

(a)

HCl

(b)

NH 4+

(c)

Cl −

(d)

H 2O2

Ans. (b)

NH 4+

contain all three types of bond

in its structure

H   |    H − N| → H    H  

+

17. Both ionic and covalent bond is present

in the following (a) (b) (c) (d)

CH4

KCl

SO 2

NaOH

Ans. (d)In O−H

NaOH

covalent bond is present in

bond while ionic bond is formed

between

OH −

and

Na +

.

18. The formation of a chemical bond is

accompanied by (a) Decrease in energy (b) Increase in energy (c) Neither increase nor decrease in energy

(d) None of these Ans.(a)Bond

formation

is

an

exothermic

reaction so there is decrease in energy of product. 19. Chemical bond implies

(a) Attraction (b)

Repulsion

(c) Neither attraction nor repulsion

(d)

Both (a) and (b)

Ans. (d) Both (a) and (b) 20. Which

of the following statements is

true (a) HF is less polar than HBr (b) Absolutely pure water does not contain any ions

(c) Chemical

bond

formation

take

place

when forces of attraction overcome the forces of repulsion (d) In covalency transference of electron takes place Ans. (c)Chemical bond formation take place when forces of attraction overcome the forces of repulsion

21. Which of the following statements is

true about

Cu (NH 3 )4 SO 4

(a) It has coordinate and covalent bonds (b) It has only coordinate bonds (c) It has only electrovalent bonds (d) It

has

electrovalent,

coordinate bonds

covalent

and

Ans. (d)It has electrovalent, covalent and coordinate bonds 22. Blue vitriol has

(a) Ionic bond (b) Coordinate bond (c) Hydrogen bond (d) All the above

Ans. (d)Blue vitriol is

.

CuSO 4 5 H 2 O

and it has all

types of bonds. 23. The

number

of

coordinate bonds in

NH 4 Cl

are respectively

(a) 1, 3 and 1 (b) 1, 3 and 2 (c) 1, 2 and 3

ionic,

covalent

and

(d) 1, 1 and 3

Ans. (a)

H  |   H − N|  H 

+   − → H  Cl  

Ionic bond = 1, Covalent bond = 3 Co-ordinate bond = 1. 24. Covalent molecules are usually held in

a crystal structure by (a) Dipole-dipole attraction

(b) Electrostatic attraction (c) Hydrogen bonds (d) Vander Waal's attraction Ans.

(d)Vander Waal's attraction

25. Strongest intermolecular hydrogen bond

is present in the following molecules pairs (a)

SiH 4

and

SiF

(b)

O || CH 3 − C − CH 3

(c)

O || H − C − OH

(d)

H2O

and

and

Ans. (c)

and

CHCl3

O || CH 3 − C − OH

H 2 O2

O || H − C − OH

and

O || CH 3 − C − OH

26. A compound contains atoms

X, Y , Z.

The

oxidation number of

X

is

+ 2, Y

is + 5 and

Z

is

−2

. Therefore, a

possible formula of the compound is

(a) (b)

X 2 (YZ 3 )2

(c)

X 3 (YZ 4 )2

(d)

X 3 (Y 4 Z )2

XYZ 2

Ans. (c) X

3

(YZ 4 )2

27. Bonds present in

CuSO 4 .5 H 2 O

is

(a) Electrovalent and covalent (b) Electrovalent and coordinate

(c) Electrovalent, covalent and coordinate (d) Covalent and coordinate Ans. (c) CuSO .5 H O has electrovalent, covalent 4

2

and coordinate bonds

Cu

2+

O   −  O − S   O 

  → O . 5 H 2O   −

28. The ionization of hydrogen atom would

give rise to (a) Hybrid ion

(b) Hydronium ion (c) Proton (d) Hydroxyl ion Ans. (c)Proton 29. Which can be described as a molecule

with residual bonding capacity (a) (b)

BeCl 2

NaCl

(c)

CH 4

(d)

N2

Ans. (a) BeCl

2

Chapter 4. Solutions Solubility 1. The solubility of a gas in water depends

on (a) Nature of the gas (b) Temperature (c) Pressure of the gas (d) All of the above Ans. (d) All of the above

2. Which of the following is not correct for

D2 O

(a) Boiling point is higher than (b) D O reacts slowly than 2

H 2O

H 2O

(c) Viscosity is higher than

H 2O

at

25 o

(d) Solubility of NaCl in it is more than

H 2O

Ans. (d) Solubility of NaCl in it is more than

H 2O

3. The statement “ The mass of a gas

dissolved in a given mass of a

solvent

at

any

temperature

is

proportional to the pressure of the gas above the solvent” is (a) Dalton’s Law of Partial Pressures (b) Law of Mass Action (c) Henry’s Law

(d) None of these Ans. (c) Henry’s Law 4. Which is correct about Henry’s law

(a) The

gas

in

contact

with

the

liquid

should behave as an ideal gas (b) There

should

not

be

any

chemical

interaction between the gas and liquid (c) The pressure applied should be high

(d) All of these Ans. (b) There should not be any chemical interaction between the gas and liquid 5. The statement “If 0.003 moles of a gas

are dissolved

in

900

g of water under a

pressure of 1 atmosphere, 0.006 moles will be

dissolved

under

atmospheres”, illustrates

a

pressure

of

2

(a) Dalton’s law of partial pressure (b) Graham’s law (c) Raoult’s law (d) Henry’s law Ans. (d) Henry’s law 6. The solution of sugar in water contains

(a) Free atoms (b) Free ions

(c) Free molecules (d) Free atom and molecules Ans. (c) Free molecules

Method of expressing concentration of solution

1.

25 ml

of

4.0 M HNO3

3.0 M HNO3

. If the

are mixed with

75 ml

volumes are additive, the

molarity of the final mixture would be (a) 3.25 M (b) 4.0 M (c) 3.75 M (d) 3.50 M Ans. (c) M V + M V 1 1

2

2

= MV

of

2. The amount of

anhydrous

Na2CO3

in 250 ml of 0.25 M solution is (a)6.225 g (b) 66.25 g (c) 6.0 g (d) 6.625 g Ans. (d) M = m wV (l) ;

0 .25 =

w 106  0 .25

; w = 6.625 gm

present

3. Dilute one litre 1 molar

H 2 SO 4

solution by

5 litre water, the normality of that solution is (a) 0 .2 N (b) 5 N (c) 10 N (d) 0.33 N Ans. (d) N V

1 1

= N 2 V2 2  1 = N 2  6 N 2 = 0.33

4. If 5.85 gms of

NaCl

are dissolved in 90

gms of water, the mole fraction of (a) 0.1 (b) 0.2 (c) 0.3 (d) 0.01 (e) 0.0196

NaCl

is

Ans. (e) 5.85 g NaCl = 90 g

H 2O =

0 .1  0 .0196 5 + 0.1

90 moles = 5 moles 18

5 .85 mole = 0 .1 mol 58 .5

mole fraction of NaCl =

.

5. The molarity of 0.006 mole of

solution is (a) 0.6 (b) 0.06

NaCl

in

100 ml

(c) 0.006 (d) 0.066 (e) None of these Ans. (b)

6.

9 .8 g

of

M=

n 0 . 006 = = 0 . 06 V (l) 0 .1

H 2 SO 4

is present in 2 litres of a

solution. The molarity of the solution is (a) 0.1M

(b) 0.05 M (c) 0.2 M (d) 0.01 M Ans. (b)

M=

W  1000 9.8  1000 = = 0.05 M mol . mass  Volume in ml . 98  2000

7. What will be the molarity of a solution

containing

5g

of sodium hydroxide in

(a) 0.5

250 ml

solution

(b) 1.0 (c) 2.0 (d) 0.1 Ans. (a)

M=

W 1000 5  1000  = = 0 .5 M m.wt. Volume in ml. 40  250

8. The normality of

(H 3 PO3 )

(a) 0.1 (b) 0.9

is

0 .3 M

phosphorus acid

(c) 0.3 (d) 0.6 Ans. (d) Basicity of Hence 0.3 M

H 3 PO3

H 3 PO3 = 0.6 N

is 2.

.

9. Which of the following has maximum

number of molecules (a) 16 gm of

O2

(b) 16 gm of

NO2

(c) 7 gm of

N2

(d) 2 gm of

H2

Ans. (d) 2 gm. Hydrogen has maximum number of molecules than others. 10. Molarity is expressed as

(a) Gram/litre

(b) Moles/litre (c) Litre/mole (d) Moles/1000 gms Ans. (b)Moles/litre 11.

20 ml

of

HCl

solution

requires

19 .85 ml

of

0.01 M NaOH

solution for complete neutralization. The molarity of

HCl

solution is

(a) 0.0099 (b) 0.099 (c) 0.99 (d) 9.9 Ans. (a) 12. How

M1V1 = M 2 V2 0.01  19 .85 = M2  20 M2 = 0.009925

much

of

;

M = 0 .0099

NaOH is required to

neutralise 1500

cm 3

of 0.1 N HCl

.

(At. wt. of Na =23)

(a) 4 g (b) 6 g (c) 40 g (d) 60 g Ans.

(b) 1500

have

number

equivalence = N1000 V 1

of NaOH

1

=

cm 3

1500  0 .1 = 0 .15 1000

= 0.15  40 = 6 gm.

of 0.1 N HCl of

gm

 0.15 gm.

equivalent

13. If 5.85 g

of NaCl (molecular weight

58.5) is dissolved in water and the solution is made up to 0.5 litre, the molarity of the solution will be (a) 0.2 (b) 0.4 (c) 1.0

(d) 0.1 w 5.85 Ans. (a) M = m.wt.  volume = = 0.2 M in litre 58 .5  0.5

14. A

mixture has 18g water and 414g

ethanol. The mole

fraction

of

water

in

mixture

(assume ideal behaviour of the mixture)

(a) 0.1

is

(b) 0.4 (c) 0.7 (d) 0.9 Ans. (a) Molecular weight of Molecular mass of 414g of 18g of

C2 H 5 OH

H 2O

has

has

=

C2 H 5 OH = 24 + 5 + 16 + 1 = 46

H 2 O = 18

414 =9 46

18 =1 18

mole

mole

Mole fraction of water

=

1 n1 1 = = 0.1 = 10 n1 + n 2 1 + 9

15. The number of molecules in 4.25 g of

ammonia is approximately (a) 0.5  10

23

(b) 1.5  10

23

(c) 3.5  10

23

(d) 2.5  10

23

Ans. (b) 17 gm NH = 1 mole. 3

Molecules of

NH 3 =

6.02  10 23  4.25 = 1.5  10 23 17

16. The largest number of molecules is in

(a) 25 g of

CO2

(b) 46 g of

C2 H 5 OH

(c) 36 g of

H 2O

(d) 54 g of

N 2 O5

Ans. (c) 36 g of

H 2O

17. If 1 M and 2.5 litre NaOH solution is

mixed with another 0.5 M and 3 litre NaOH solution, then molarity of the resultant solution will be (a) 1.0 M (b) 0.73 M (c) 0.80 M

(d) 0.50 M Ans. (b) (2.5 1 + 3  0.5) = M or

2. 5 + 1. 5 = M 3  5. 5

18. When

a

or

3

 5. 5

M3 =

4 = 0 .73 5 .5

solute

is

M. present

quantities the following expression is used (a) Gram per million (b) Milligram percent

in

trace

(c) Microgram percent (d) Nano gram percent (e) Parts per million Ans. (e)Parts per million 19. When the concentration is expressed as

the number of moles of a solute per litre of solution it known as

(a)

Normality

(b)

Molarity (c) Mole fraction (d) Mass percentage (e) Molality Ans. (b)Molarity 20.

The normality of 2.3 M

H 2 SO 4

solution is

(a) 2.3 N (b) 4.6 N (c) 0.46 N (d) 0.23 N Ans. (b) Normality of

= 2 .3  2 = 4 .6 N

2.3 M H 2 SO 4 = M 

Valency

21. The

molarity of a solution made by

mixing 50ml of conc.

(a)

36 M

(b)

18 M (c) 9 M (d) 6 M

H 2 SO 4

(36N) with 50 ml of water is

Ans. (c) N V

1 1

N2 =

36  50 = 18 100

22.

;

171

= N 2 V2

,

36  50 = N 2  100

18 N H 2 SO 4 = 9 M H 2 SO 4

g

of

.

cane

sugar

(C12 H 22O11 )

dissolved in 1 litre of water. The molarity of the solution is (a)

2.0 M

(b)

1.0 M (c) 0.5 M

is

(d) 0.25 M Ans. (c)Molarity

23.

The

=

171 w = = 0.5 M 342  1 m.wt.  volume in litre

volumes

of

4 N HCl

and

.

10 N HCl

required to make 1 litre of

6 N HCl

are

(a) 0.75 litre of 10 N HCl and 0.25 litre of 4 N HCl

(b) 0.25 litre of 4 N HCl and

0.75 litre of 10 N HCl

(c) 0.67 litre of 4 N HCl and 0.33 litre of 10 N HCl (d) 0.80 litre of 4 N HCl and 0.20 litre of 10 N HCl (e) 0.50 litre of 4 N HCl and 0.50 litre of 10 N HCl Ans. (c) N V + N V 1 1

4 x + 10 (1 − x ) = 6  1

2

;

2

= NV

−6 x = −4

;

x = 0 .66

24.

Which statement is true for solution of

0.020 M

H 2 SO 4

(a) 2 litre of the solution contains 0.020

mole of

SO 42 −

(b) 2 litre of the solution contains 0.080

mole of

H 3O +

(c) 1 litre of the solution contains 0.020

mole

H 3O +

(d) None of these Ans. (b) [H O 3

+

] = 2  0 . 02 = 0 . 04 M

 2 litre solution contains 0.08 mole of H 3O +

.

25.10 litre solution of urea contains 240g

urea. The active mass of urea will be (a) 0.04

(b) 0.02 (c) 0.4 (d) 0.2 Ans. (c)10 litre of urea solution contains 240 gm of urea  Active mass = 60240 10 = 0.4 .

26.

5 ml of N HCl, 20 ml of N/2

30 ml of

H 2 SO 4

and

N/3 HNO3 are mixed together and volume made to one litre. The normally of the resulting solution is (a)

N 5

(b) 10N (c)

N 20

(d) 40N (e) 25N

Ans. (d) NV = N V + N V 1 1

or,

1000 N = 1  5 +

27. The

2

2

+ N 3 V3

1 1  20 +  30 2 3

or

amount of

N=

1 40

.

K2 Cr2 O7

(eq. wt. 49.04)

required to prepare 100 ml of its 0.05 N solution is (a) 2.9424 g (b) 0.4904 g (c) 1.4712 g

(d) 0.2452 g .wt.  V (ml ) 0 .05  49 .04  100 = Ans. (d) W = N  eq1000 1000

28.

= 0 .2452 .

With increase of temperature, which of

these changes (a) Molality (b) Weight fraction of solute (c) Fraction of solute present in water

(d) Mole fraction Ans. (c) Fraction of solute present in water 29.

25ml of a solution of barium hydroxide

on titration

with

a

0.1molar

solution

of

hydrochloric acid gave a litre value of 35

ml.

The

molarity

solution was (a) 0.07

of

barium

hydroxide

(b) 0.14 (c) 0.28 (d) 0.35 Ans. (a)For HCl

N 1 V1 = N 2 V2

N1 =

0 .1  35 25

30.

;

;

M =

M = N = 0 .1

25  N 1 = 0.1  35

0 .1  35 = 0 .07 25  2

.

2.0 molar solution is obtained , when

0.5 mole

solute is dissolved in (a)250 ml solvent (b) 250 g solvent (c) 250 ml solution (d) 1000 ml solvent Ans. (c)We know that Molarity

 2.0 =

=

Number of moles of solute Volume of solution in litre

0 .5 Volume of solution in litre

 Volume of solution in litre

=

0 .5 = 0 .250 litre = 250 ml . 2.0

31. How many gram of HCl will be present

in

150 ml

of

its 0.52 M (a) 2.84 gm (b) 5.70 gm (c) 8.50 gm

solution

(d) 3.65 gm Ans. (a) M = m wV (l) ;

32.

0 .52 =

is

(a) 0.5 (b) 0.1 (c) 1

;

w = 2.84 gm

The number of moles present in 2

litre of 0.5 M

NaOH

w 36 .5  0 .15

(d) 2 Ans. (c)

33.

36g

M =

n V (l)

;

0 .5 =

water

n 2

;

and

n =1

828g

ethyl

alcohol

form an ideal solution. The mole fraction of water in it, is (a)1.0 (b) 0.7 (c) 0.4

(d) 0.1 Ans. (d)

x H 2O =

34.

N=

W 828 w 36 = = 18 , n = = =2 M 46 m 18

n 2 2 = = = 0.1 2 + 18 20 n+N

What

will

be

the

normality

of

a

solution containing 4.9 g. H PO 3

water (a) 0.3

4

dissolved in 500 ml

(b) 1.0 (c) 3.0 (d) 0.1 Ans. (a) 0.3 w  1000 (a) N = E  volume , in ml .

N =

4 .9  1000 = 0.3 N 32 .6  500

E=

98 = 32 .6 3

.

35.3.0 molal NaOH

of 1.110

solution has a density

g/ml. The molarity of the solution is (a) 3.0504 (b) 3.64 (c) 3.05 (d) 2.9732 Ans. (d)2.9732 36.

Which

of

expressing

the

following

modes

of

concentration is independent of temperature

(a) Molarity (b) Molality (c) Formality (d) Normality Ans. (b)Molality 37. The molality of a solution is

(a) Number of moles of solute per

1000 ml

of

the solvent (b) Number of moles of solute per

1000 gm

of

the solvent (c) Number of moles of solute per

1000 ml

of

the solution (d) Number of gram equivalents of solute per

1000 ml

the solution

of

Ans. (b)Number of moles of solute per

1000 gm

38.

of the solvent The number of molecules in

methane is (a) 3.0  10

23

(b) 6.02  10 (c)

23

16  10 23 6 .02

(d) 316.0  10

23

16 gm

of

Ans. (b) 6.02  10 39.

23

The number of moles of a solute in

its solution is 20 and total number of moles are 80. The mole fraction of solute is (a) 2.5 (b) 0.25 (c) 1

(d) 0.75 Ans. (b)Mole fraction of solute

=

20 = 0.25 80

.

40. The normality of a solution of sodium

hydroxide 100 ml of which contains 4 grams of (a)0.1 (b) 40 (c) 1.0

NaOH

is

(d) 0.4  1000 4  1000 = = 1 .0 N . Ans. (c) N = m.wt. wVolume in ml 40  100

41. Two

solutions

of

a

substance

(non

electrolyte) are mixed in the following manner 480 ml of 1.5M first solution + 520 mL of 1.2M second solution. What is the molarity of the final mixture (a) 1.20 M

(b) 1.50 M (c) 1.344 M (d) 2.70 M Ans. (c) M V + M V 1 1

2

2

= M 3 V3

;

1 .5  480 + 1 .2  520 = M  1000

M=

42.

720 + 624 1000

= 1.344 M

.

The normal amount of glucose in

of blood

100 ml

(8–12 hours after a meal) is (a) 8 mg (b) 80 mg (c)

200 mg

(d) 800 mg Ans. (b) 80 mg 43.

Molar solution means 1 mole of solute

present

in (a) 1000g of solvent (b) 1 litre of solvent (c) 1 litre of solution (d) 1000g of solution Ans. (c)1 litre of solution 44.

What will be the molality of a solution

having

18 g

500 g

of glucose (mol. wt. = 180) dissolved in of water

(a) 1m (b) 0.5 m (c) 0.2 m (d) 2 m 18  1000 = 0 .2 m Ans. (c) m = 180  500

45.

A

solution

of

Al2 (SO 4 )3 {d = 1.253 gm / ml }

contain

22% salt by

weight.

The

molarity,

molality of the solution is (a) 0.805 M, 4.83 N, 0.825 M (b) 0.825 M, 48.3 N, 0.805 M (c) 4.83 M, 4.83 N, 4.83 M (d) None

normality

and

Ans. (a)Molarity Normality Molality

46.

=

=

=

%  10  d GMM

%  10  d GEM

=

=

22  10  1 .253 = 0 .805 M 342

.

22  10  1 .253 = 4 .83 N 342 / 6

22  1000 = 0 . 825 m 342 (100 − 22 )

Which of the following should be done

in order to

prepare

0.30 M NaCl

(a) Add

0.40 M NaCl

(mol.wt. of

0.585 g NaCl

starting NaCl = 58 .5

)

with

100 ml

of

(b) Add (c) Add

20 ml

water

0.010 ml NaCl

(d) Evaporate

10 ml

water

Ans. (a)100 ml. of 0.30M = 1001000 0.3 = 0.03 mole of

NaCl 100 ml of 0.40M

NaCl

=

100  0 .4 = 0 .04 1000

mole of

Moles

of

NaCl

to

be

added

= 0.04 − 0.03 = 0.01 mole

= 0.585 gm 47. Which of the following solutions has the

highest normality (a) 8 gm of

KOH

/ litre

(b) N phosphoric acid

(c) 6 gm of (d) 0.5 M H SO 2

NaOH

/ 100 ml

4

Ans. (c)

N=

6  1000 = 1.5 N 40  100

It is show highest normality than others. 48.

What volume of

0.1 mole of the solute (a) 100 ml

0. 8 M

solution contains

(b) 125 ml (c) 500 ml (d) 62 .5 ml Ans. (b) M = Vn(l) 

49.

0 .8 =

0 .1  V = 125 ml V (l)

.

Hydrochloric acid solution

have

A

and

B

concentration The

of

volumes

0 .5 N

of

required to make (a) 0.5 l of

A + 1.5 l

of

B

(b) 1.5 l of

A + 0 .5 l

of

B

(c) 1.0 l of

A + 1.0 l

of

B

(d) 0.75 l of

A + 1 .25 l

Ans. (a) 0.5 l of

of

A + 1.5 l

and

0 .1 N

solutions

2 litres

of

B

of

B

0 .2 N

respectively.

A

HCl

are

and

B

50.

Conc.

H 2 SO 4

has

a density of 1.98

gm/ml and is 98%

H 2 SO 4

(a)

2 N

(b)

19.8 N

by weight. Its normality is

(c) 39.6 N (d) 98 N Ans. (c)

Strength of

H 2 SO 4 = 98  19 .8 g / litre

S = eq. wt.  N

;

N=

S 98  19 .8 = = 39 .6 eq. wt. 49

51. The mole fraction of the solute in one

molal aqueous solution is (a) 0.027 (b) 0.036 (c) 0.018 (d) 0.009

Ans. (c)

N=

x Solute

=

W = 1000 gm

(H 2 O)

;

n =1

mole

W 1000 = = 55 .55 M 18

n 1 = n + N 1 + 55 .55

= 0.018.

52.With 63 gm of oxalic acid how many

litres of

N 10

solution can be prepared (a)100 litre (b) 10 litre

(c) 1 litre (d) 1000 litre Ans. (b)10 litre 53.Molarity of

(a) 0.2 (b) 0.4 (c) 0.6 (d) 0.1

0.2 N H 2 SO 4

is

Ans. (d)

Normality

of

acid

=

substance

of

molarity  basicity

i.e., 0.2=molarity  2  Molarity = 0.2/2 = 0.1 10.6

54.

grams

of

a

molecular weight 106

was

dissolved

in

100 ml

.

10 ml

solution was pipetted out into a

of 1000 ml

this flask

and made up to the mark with distilled

water. The molarity of the resulting solution is (a) 1.0M (b) 10

−2

(c) 10

−3

(d) 10

−4

Ans.

M

M

M

(b) 10

−2

M

55. The

mole

fraction

of

water

in

20%

aqueous solution of

H 2 O2

is

(a) 7768 (b) 7768 (c)

20 80

(d) 8020

Ans. (b)Mole fraction of

H 2O

=

80 18 80 20 + 18 34

=

68 77

.

56.

Mole fraction

(X )

of any solution is

equal to (a)

No. of moles of solute Volume of solution in litre

of gram equivalent of solute (b) No.Volume of solution in litre

(c)

No. of moles of solute Mass of solvent in kg

(d)

No. of moles of any constituent Total no. of moles of all constituents

No. of moles of any constituent Ans. (d) Total no. of moles of all constituents

57. When

WB gm

solute (molecular mass

MB

)

dissolves in solvent. The molality

WA gm

is (a) WW

B A



MB 1000

(b)

WB 1000  MB WA

(c)

W A 1000  WB MB

M (d) WW 1000 A

B

B

M

of the solution

Ans. (b) MW

58.

(a)

B B



1000 WA

Normality

(N )

of a solution is equal to

No. of moles of solute Volume of solution in litre

of gram equivalent of solute (b) No.Volume of solution in litre

(c)

No. of moles of solute Mass of solvent in kg

(d) None of these of gram equivalent of solute Ans. (b) No.Volume of solution in litre

The volume strength of

59.

1. 5 N H 2 O 2

is (a) 4.8 (b) 5.2 (c) 8.8 (d) 8.4 Ans. (d)Volume strength

=

1.5  100 = 8.82 . 17

solution

60. How many

0 .25 gm

gm

of

H 2 SO 4

mole of

H 2 SO 4

(a) 24.5 (b) 2.45 (c) 0.25 (d) 0.245 Ans. (a) n = mw ;w = n  m = 0.25  98

= 24 .5 gm

is present in

61. 20 g of hydrogen is present in 5 litre

vessel. The molar concentration of hydrogen is (a) 4 (b) 1 (c) 3 (d) 2 Ans. (d)Molar concentration

[H 2 ] =

20 2 Mole = =2 V in litre 5

.

62.

To prepare a solution of concentration

of 0.03

g/ml of

AgNO3

, what amount of

be added in 60 ml of solution (a) 1.8 (b) 0.8 (c) 0.18 (d) None of these

AgNO3

should

Ans. (a) Amount of solution 63.

AgNO3

added in 60 ml of

= 60  0.03 = 1.8 g

How many grams of dibasic acid (mol.

wt. 200) should be present in

100 ml

of its aqueous

solution to give decinormal strength (a) 1g (b) 2 g (c) 10 g

(d) 20 g Ans. (a) N = E wV (l)  0.1 = 100 w 0.1  w = 1gm The weight of pure

64.

prepare

250 cm 3

(a) 4 g (b) 1g (c)

2g

of

0 .1 N

solution is

NaOH

required to

(d) 10 g Ans. (b) N = E wV(l) 

65.

If

20 ml

of

0.1 =

w  w = 1 gm 40  0 .25

0.4 N NaOH

solution completely

neutralises

40 ml

of a dibasic acid. The molarity of the

acid solution is (a) 0.1M (b) 0.2 M

(c) 0.3 M (d) 0.4 M Ans. (a)

66.

20  0 .4 = 40  N or N = 0 .2 or M =

0.2 = 0. 1 M 2

.

Which of the following concentration

factor is affected by change in temperature (a) Molarity (b) Molality

(c) Mole fraction (d) Weight fraction Ans. (a)Molarity 67. The distribution law is applied for the

distribution of basic acid between (a) Water and ethyl alcohol (b) Water and amyl alcohol

(c) Water and sulphuric acid (d) Water and liquor ammonia Ans. (c)Water and sulphuric acid 68.

Which is heaviest

(a) 25 gm of mercury (b) 2 moles of water (c) 2 moles of carbon

dioxide

(d) 4 gm atoms of oxygen

Ans. (c)2 moles of carbon

The molarity of a solution of

69.

having

10 .6 g / 500 ml

(a) 0.2 M (b) 2 M (c)

dioxide

20 M

(d) 0.02 M

of solution is

Na2 CO3

Ans.

On passing

70.

of

Zn +2

 1000 (a) M = m.wt. wVolume in ml .

Cu +

H2S

=

10 .6  1000 = 0 .2 M 106  500

.

gas through a solution

and

ions,

CuS is precipitated first because (a) Solubility product of CuS is equal to the ionic product of ZnS

(b) Solubility product of CuS is equal to the solubility product of ZnS (c) Solubility product of CuS is lower than the solubility product of ZnS (d) Solubility product of CuS is greater than the solubility product of ZnS Ans.

(d)Solubility product of CuS is

greater than the solubility product of ZnS

71. The number of moles of solute per kg

of a solvent is called its (a) Molarity (b) Normality (c) Molar fraction (d) Molality Ans. (d)Molality

72. 1.0 gm of pure calcium carbonate was

found to require 50 ml of dilute

HCl

reaction. The strength of the given by (a) 4 N (b) 2 N (c) 0.4 N (d) 0.2 N

for complete

HCl

solution is

M.eq. of HCl = M.eq. of

Ans. (c)

CaCO3

N  50 =

1  1000 50

;

N=

1  1000 = 0.4 N 50  50

73. Molecular weight of glucose is 180. A

solution of glucose which contains 18 gms per litre is

(a) 2 molal

(b) 1 molal (c) 0.1 molal (d) 18 molal 18 = 0 .1 molal . Ans. (c) molality = 180

74. 0.5 M of

H 2 SO 4

is diluted from 1 litre to

10 litre, normality of resulting solution is (a)1 N

(b) 0.1 N (c) 10 N (d) 11 N Ans. (b)Molarity of Normality of

H 2 SO 4 = 0.5

H 2 SO4 (N1 ) = 0.5  2 = 1

N1V1 = N 2 V2

1  1 = N 2  10

or

N2 =

1 = 0 .1 N . 10

75. If one mole of a substance is present in

1 kg

of

solvent, then (a) It shows molar concentration (b) It shows molal concentration (c) It shows normality (d) It shows strength

gm / gm

Ans. (b) It shows molal concentration

76. The molality of 90%

H2 SO4

solution is

[density=1.8 gm/ml] (a)1.8 (b) 48.4 (c) 9.18 (d) 94.6 Ans. (c)

= 1.8 gm / ml

The density of solution

Weight of one litre of solution



Weight

of H SO 2

4

= 1800 gm

in

the

solution = 1800100 90 =1620gm





Weight of solvent

Molality

=

= 1800 − 1620 = 180 gm

1620 100  = 9 .18 98 180

77. The volume of water to be added to

of 0.5 N

H 2 SO 4

to get decinormal concentration is

100 cm 3

(a)400 (b)

500

cm 3

cm 3

(c) 450

cm 3

(d) 100

cm 3

Ans. (a) of water = x

100 cm 3  0.5 N = x  0.1 N

x =

100  0 .5 = 500 cm 3 0 .1

Suppose the total volume

Therefore the volume of water added

= Total volume – 100cm

3

= 500 − 100 = 400 cm 3

.

78.If 25 ml of 0.25 M NaCl solution is

diluted with water to a volume of 500ml the new concentration of the solution is (a)

0.167 M

(b)

0.0125 M

(c) 0.833 M (d) 0.0167 M Ans. (b) M V

1 1

= M 2 V2

,

M2 =

0 .25  25 = 0.0125 500

.

79. 10 grams of a solute is dissolved in 90

grams of a solvent. Its mass percent in solution is (a) 0.01 (b) 11.1

(c) 10 (d) 9 Ans. (c)

=

% by

wt. =

wt. of the solute (g)  100 wt. of the solution g

10  100 = 10 90 + 10

80. What

is the molality of a solution

which contains 18 g of glucose water

(C6 H12O6 )

in 250 g of

(a)4.0 m (b) 0.4 m (c) 4.2 m (d) 0.8 m Ans. (b)Molality

=

w 18  1000  1000 = = 0.4 m W 180  250

m

81. Calculate the molality of 1 litre solution

of 93%

H 2 SO 4

(weight/volume).

solution is

The

density

1.84 g /ml

(a) 10.43 (b) 20.36 (c) 12.05 (d) 14.05 Ans. (d)Molality

(m ) =

w  1000 = 14 .05 mW

.

of

the

82.

Volume of water needed to mix with

10 ml 10N

HNO3

to get 0.1 N

(a)

1000 ml

(b)

990 ml (c) 1010 ml (d) 10 ml Ans. (b) N V

1 1

= N 2 V2

HNO3

10  10 = 0.1 (10 + V)

V=

10  10 − 10 = 1000 − 10 = 990 ml . 0.1

83.

The sum of the mole fraction of the

components of a solution is (a) 0 (b) 1 (c) 2

(d) 4 Ans. (b)Sum of mole fraction is always 1. 84.

Increasing

the

temperature

aqueous solution will cause (a) Decrease in molality (b) Decrease in molarity (c) Decrease in mole fraction

of

an

(d) Decrease in % w/w Ans.

(b)An

increases

in

temperature

increase the volume of the solution and thus decreases its molarity. 85.

1000 gms aqueous solution of

CaCO3

contains 10 gms of solution is (a) 10 ppm

carbonate. Concentration of the

(b) 100 ppm (c) 1000 ppm (d) 10000 ppm Ans. (d) 10 parts of 3

CaCO3

has number of parts

= 10

10 6

=

parts of

10  10 6 = 10 ,000 ppm 10 3

.

CaCO3

has number of parts

86.

3.65 gms of HCl is dissolved in 16.2

gms of water. The mole fraction of HCl in the resulting solution is (a) 0.4 (b) 0.3 (c) 0.2 (d) 0.1

Ans. (d) X = n +n N

n=

w 3 .65 W 16 .2 = = 0 .1, N = = = 0 .9 m 36 .5 M 18

X = 0.10+.10.9 = 0.1. 87.An aqueous solution of glucose is 10%

in strength. The volume in which dissolved will be (a) 18 litre

1 gm

mole of it is

(b) 9 litre (c) 0.9 litre (d) 1.8 litre Ans.

(d)10% 10 g =

10 mole 180

in

glucose

solution

means

100 cc. i.e., 0.1 litre

Hence 1 mole will be present in 0.110 180 =1.8

litre.

88. The

concentration

of

an

aqueous

solution of

0.01 M CH3OH

solution

is

very

which of the following (a)

0.01 % CH 3OH

(b) 0.01m CH OH 3

(c)

x CH 3 OH = 0.01

(d) 0.99 M H O 2

nearly

equal

to

(e) 0.01 N CH OH 3

Ans. (e)For methyl alcohol N = M. 89.

When

H 2O

1.80 gm

glucose dissolve in

, the

mole fraction of glucose is (a) 0.00399 (b) 0.00199 (c) 0.0199

90 gm

of

(d) 0.998 Ans. (b)Mole fraction of glucose = =

n n+N

0 .01 = 0 .00199 0 .01 + 5

90.

6 .02  10 20

molecules of urea are present

in 100 ml of its solution. The concentration of urea solution is (a) 0.02 M

(b) 0.01 M (c) 0.001 M (d) 0.1 M (Avogadro constant, Ans. (b)Mole of urea

N A = 6 .02  10 23 mol −1 )

=

6 .02  10 20 = 10 − 3 6.02  10 23

moles

Conc. of solution (in molarity) =

10 −3  1000 = 0.01 M 100

.

91. The number of moles of

SO 2 Cl2

in

13 .5 gm

is

(a) 0.1 (b) 0.2 (c) 0.3 (d) 0.4 Ans. (a) 135

Gram molecule of

SO 2 Cl2

=

n=

w 13 .5 = = 0 .1 m 135

.

The

92.

prepare of

0 .2 N

(a) 126 g (b) 12 .6 g (c) 63 g (d) 6.3 g

weight 500 ml

solution is

of

H 2C2O4 . 2 H 2O

required

to

Ans. (d)1000 ml of 1 N oxalic solution = 63 g 500 ml of 0.2 N oxalic acid solution =

93.

63  500  0 .2 = 6 .3 g 1000

.

In a solution of

7.8 gm

benzene

C6 H 6

and

46 .0 gm

toluene

(C6 H 5 CH 3 )

,

the

mole

benzene in this solution is

fraction

of

(a) 1 / 6 (b) 1 / 5 (c) 1 / 2 (d) 1 / 3 Ans. (a)Mole fraction at

94.

A solution contains

50 % CH 3 COOH

would be

7 .8 1 78 C6 H 6 = = 7 .8 46 6 + 78 92

25 %H 2 O

.

, 25 %C H OH and 2

5

by mass. The mole fraction of

H 2O

(a) 0.25 (b) 2.5 (c) 0.503 (d) 5.03 Ans. (c) X

95.

H 2O

=

nH 2 O n H 2 O + nC 2 H 5 OH + nCH 3 COOH

A 5 molar solution of from 1 litre

H 2 SO 4

is diluted

to 10 litres. What is the normality of the solution (a) 0.25 N (b) 1 N (c) 2 N (d) 7 N Ans. (a)

i.e.

M1V1 = M 2 V2

5  1 = M2  10  M2 = 0.5

Normality of the solution

96.

=

0 .5 = 0 .25 . 2

Molarity of a solution containing

in

250 ml

of solution is (a) 0.1M (b) 1 M (c) 0.01 M (d) 0.001 M

1g NaOH

Ans. (a)

M=

1  1000 w  1000 = = 0.1 M . 40  250 m  Volume in ml .

97. What is molarity of a solution of HCl

which contains 49% by weight of solute and whose specific gravity is 1.41 (a)

15.25

(b)

16.75 (c) 18.92

(d) 20.08 Ans. (c)18.92 98.

NaClO

solution reacts with

as,

H 2 SO 3

NaClO + H 2 SO 3 → NaCl + H 2 SO 4

. A

solution of

NaClO

used in the above reaction

contained

15g

of

NaClO

per

litre.

normality of the solution would be (a) 0.8

The

(b) 0.6 (c) 0.2 (d) 0.33  1000 Ans. (d) N = eq. wt. w volume = 0 .33 N . in ml .

99.

A solution contains

1.2046  10 24

hydrochloric

acid molecules in one

dm 3

of the solution. The

strength of the solution is

(a) 6 N (b) 2 N (c) 4 N (d) 8 N Ans.

(b)Mole of

HCl =

1.2046  10 24 = 2 mole 6.023  10 23

Normality = molarity  basidity or acicity

= 2 1 = 2N

100.

10 N

and

1 N 10

solution is called

(a) Decinormal and decanormal solution (b) Normal and decinormal solution (c) Normal and decanormal solution (d) Decanormal and decinormal solution Ans. (d) 10 N = Deca - normal , 101. When

7.1gm Na 2 SO 4

1 10

N = Deci-normal. (molecular mass 142)

dissolves in

100 ml H 2 O

, the molarity of the solution is

(a) 2.0 M (b) 1.0 M (c) 0.5 M (d) 0.05 M Ans. (c)Molarity

=

w  1000 ml wt.  Volume ml .

102. Molarity of 4%

(a) 0.1M (b) 0.5 M

NaOH

.1  1000 = 7142  100

= 0 .5 M

solution is

.

(c) 0.01 M (d) 1.0 M Ans. (d) M = 4 40 10 = 1 M . 103. When

6 gm

urea dissolve in

mole fraction of urea is (a) 1010.1 (b) 1010.1

180 gm H 2 O

. The

(c)

10 . 1 0 .1

(d) 100..11 Ans.

(d)Mole

6 n 0 .1 60 X= = = 6 180 n+N 10 .1 + 60 18

fraction

.

104. The normality of 10% (weight/volume)

acetic acid is (a) 1 N

(b) 10 N (c) 1.7 N (d) 0.83 N Ans. (c) N = Eq .wtw .1000 Volume

=

10  1000 = 1 . 66 N 60  100

.

105. Unit of mole fraction is

(a) Moles/litre (b) Moles/litre2 (c) Moles–litre

(d) Dimensionless Ans. (d)Dimensionless 106. Normality of

2M

sulphuric acid is

(a) 2 N (b) 4 N (c)

N /2

(d) N / 4 Ans. (b) N = M  bosicity ;

N = 22 = 4

.

107. Molar concentration

= (a)

No. of moles of solute Volume of solution in litre

of gram equivalent of solute (b) No.Volume of solution in litre

(c)

No. of moles of solute Mass of solvent in kg

(d)

No. of moles of any constituent Total no. of moles of all constituents

No. of moles of solute Ans. (a) Volume of solution in litre

(M )

of any solution

108. If

5 .0 gm

of

BaCl2

is

present

in

solution, the concentration is (a) 1 ppm (b) 5 ppm (c) 50 ppm (d) 1000 ppm Ans. (b)Concentration

=

5  10 6 10 6

= 5 ppm.

10 6 gm

109. 1 Molar solution contains

(a) 1000g of solute (b) 1000g of solvent (c) 1 litre of solvent (d) 1 litre of solution Ans. (d)1 litre of solution 110. To neutralise completely 20 mL of 0.1

M

aqueous solution of phosphorous acid

(H 3 PO3 ),

the volume of

0.1 M aqueous KOH solution required is

(a) 40 mL (b) 20 mL (c) 10 mL (d) 60 mL

Ans. (a)

N 1 V1

H 3 PO3

(acid)

is a dibasic acid

= N 2 V2

(base)

0.1  2  20 = 0.1  1  V2

 V2 =

111.

0 .1  2  20 = 40 ml 0 .1  1

On dissolving 1 mole of each of the

following acids in 1 litre water, the acid which does not give a solution of strength

1N

is

(a)

HCl

(b) Perchloric acid (c)

HNO3

(d) Phosphoric acid Ans.

(d) H PO ⇌ 3

4

H + + H 2 PO 4−

H 2 PO 4−

⇌

H + + HPO 42 −

HPO 42 −

⇌

H + + PO 43 −

Phosphoric acid does not give 1N strength.

112. How

required

many to

grams

neutralize

of 12.2

benzoic acid (a)

40 gms

(b)

4 gms

(c)

16 gms

(d) 12 .2 gms Ans. (b) C H COOH + NaOH →C H COONa + H O 6

5

6

5

2

will

be

grams

of

NaOH

w 12 .2 = = 4 gms . 40 122

113.

10 ml

of conc.

H 2 SO 4

(18 molar) is diluted

to 1 litre. The approximate strength of dilute acid could be (a) 0.18 N (b) 0.09 N (c) 0.36 N (d) 1800 N

Ans. (c)

(H 2 SO 4 ) N 1 V1

=

N 2 V2

(dilute

acid)

N 2 = (10  36 ) / 1000 = 0.36 N

114. The

.

normality

hydrogen peroxide is (a)

0.176

(b)

3.52 (c) 1.78

of

10

lit.

volume

(d) 0.88 (e) 17.8 Ans. (c) 1 M

H 2O2 → H 2O +

H 2 O2

1 O2 2

solution

= 2 N = 34 gm / litre = 11 .2

So

Normality = 211 10.2 = 1.75 115. Essential

quantity

of

ammonium

sulphate taken for preparation of 1 molar solution in 2 litres is

(a) (b) (c) (d)

132 gm

264 gm

198 gm

212 gm

Ans. (b)Weight = molarity × m.wt.× v =

1  132  2 = 264 gm.

116. In a mixture

of 1 gm

H2

and 8 gm

the mole fraction of hydrogen is

O2

,

(a) 0.667 (b) 0.5 (c) 0.33 (d) None of these Ans. (a)Mole fraction

117. A solution of

w n m = = w W n+N + m M

CaCl2

is

moles of chloride ion in (a) 0.25

500 ml

=

0.5 mol / litre

1 2

1 8 + 2 32

= 0 .667

.

, then the

will be

(b) 0.50 (c) 0.75 (d) 1.00 Ans. (b)0.50 118. What is the molarity of

H 2 SO 4

that has a density 1.84 gm/cc at contains solute 98% by weight (a) 4.18 M

solution, 35 o C

and

(b) 8.14 M (c) 18.4 M (d) 18 M Ans. (c) 98%

H 2 SO 4

means 98g

solution. 100 cc = 54 .3 cc 1 .84

;

Hence molarity

98 g H 2 SO 4 = 1 mol

=

1  1000 = 18 .4 M 54 .3

H 2 SO 4

in 100g

119. A

certain

aqueous

solution

of

(formula mass =162) has a density of and contains

20 .0% FeCl3

this solution is (a) 0.028 (b) 0.163 (c) 1.27 (d) 1.47 Ans.

(c)

1.27

FeCl3

1.1 g / ml

. Molar concentration of

120. If 0.50 mol of

mol of

Na3 PO4

moles of

Ca3 (PO4 )2

CaCl2

is mixed with 0.20

, the maximum

which can be formed, is

(a) 0.70 (b) 0.50 (c) 0.20 (d) 0.10 Ans. (d)

number of

3 CaCl2 + 2 Na3 PO4 → Ca3 (PO4 )2 + 6 NaCl



Mole of

Na3 PO4 = 3

mole of

CaCl2 = 1

mole

Ca3 (PO4 )2



0.2 mole of

mole of 121. An

Ca3 (PO4 )2

X

Na3 PO4 = 0.3

mole of

CaCl2

= 0.1

.

molal solution of a compound in

benzene has mole fraction of solute equal to 0.2. The value of (a) 14 (b) 3.2

X

is

(c) 4 (d) 2 Ans. (b)

X = 0 .2 1000 X+ 78

122. Molecular

weight of urea is 60. A

solution of urea containing litre is (a) 1 molar (b) 1.5 molar

6g

urea in one

(c) 0.1 molar (d) 0.01 molar Ans. (c) C = 606 = 0.1 molar. 123. The molar solution of sulphuric acid is

equal to (a) N solution (b)

2N

solution

(c)

N /2

solution

(d)

3N

solution

Ans. (b)Molar solution of sulphuric acid is equal to 2N because it is show dibasic nature. 124. The

weight

of

sodium

required to prepare 500 ml normal solution is (a) 13.25 g (b) 26.5 g

carbonate of a semi-

(c) 53 g (d) 6.125 g Ans. (a)

w=

N=

0 .5  53  500 = 13 .25 1000

125.

200 ml

eq . wt. =

106 = 53 2

.

of

dissolved concentration

(Na = 23 ; Cl = 35 .5)

w  1000 eq. wt.  volume in ml .

a

solution

sodium of

the

contains

chloride. solution

5.85 g

The will

be

(a) 1 molar (b) 2 molar (c) 0.5 molar (d) 0.25 molar Ans. (c)Molar concentration = 126. Molarity

.

of a solution prepared by

dissolving 75.5 g of pure KOH solution is

5 .85  1000 = 0 .5 Molar 58 .5  200

in 540 ml

(a) 3.05 M (b) 1.35 M (c) 2.50 M (d) 4.50 M 75 .5  1000 Ans. (c) M = m.wtw .1000 = V in ml 56  540

127. Which

= 2 .50

M

one of the following is an

extensive property (a) Molar volume

(b) Molarity (c) Number of moles (d) Mole fraction Ans. (c) Number of moles 128. Addition of conc. HCl to saturated

solution precipitates

BaCl2

; because

(a) It follows from Le Chatelier’s principle (b) Of common-ion effect

BaCl2

(c) Ionic product

(Ba ++ ), (Cl − )

remains constant in

a saturated solution (d) At constant temperature, the product (Cl − )2

remains

constant

in

a

(Ba 2+ ),

saturated

solution Ans.

(c)Ionic

product

(Ba ++ ), (Cl − )

remains

constant in a saturated solution 129. How much water is needed to dilute

10 ml of 10 N

hydrochloric

acid

to

make

it

exactly

decinormal (0.1 N) (a) 990 ml (b) 1000 ml (c) 1010 ml (d) 100 ml Ans. (a)

10  10 = 0 . 1

N 1 V1 = N 2 V2

× Volume of new solution

Volume of water = 1000 – 10 = 990 ml. 130. The formula weight of

weight of the acid in is (a) (b) (c) (d)

2.45 g

3.92 g

4.90 g

9. 8 g

400 ml

H2 SO4

of

is 98. The

0 .1 M

solution

m. w t.  V Ans. (b) W = M  1000

=

0 .1  98  400 = 3 .92 g 1000

.

131. The molarity of pure water is

(a) 55.6 (b) 5.56 (c) 100 (d) 18 Ans. (a)Molarity of pure water

=

1000 = 55 .6 M 18

.

132. The molarity of a

be (a) 0.05 M (b) 0.2 M (c) 0.1 M (d) 0.4 M Ans. (c) M = N2 = 02.2 = 0.1 M

0.2 N Na2 CO3

solution will

133. How many moles of water are present

in 180

g

of water

(a) 1 mole (b) 18 mole (c) 10 mole (d) 100 mole Ans. (c)Moles of water

=

180 = 10 mole . 18

134. If we take

44 g

of

CO2

will be mole fraction of

and

CO2

14 g

of

what

N2

in the mixture

(a) 1/5 (b) 1/3 (c) 2/3 (d) ¼ Ans. (c) Mole fraction of CO2 = n

n CO 2 CO 2

+ n N2

=

44 44 44 14 + 44 28

=

2 3

.

135. What is the volume of

to react completely with carbonate

(Ca = 40 , C = 12

(a) 150 cm

3

(b)

250 cm 3

(c)

200 cm 3

(d) 100 cm

3

and

O = 16 )

1.0 g

0.1 N HCl

required

of pure calcium

Ans. (c) 200 cm

3

136. The amount of

a

0.100 M NaOH

NaOH

in gms in

solution would be

(a) 4 gm (b) 2 gm (c) 1 gm (d) 2.5 gm Ans. (c) M = m wV (l) 

0.1 =

w4  w = 1 gm 40  1

250 cm 3

of

137. 4.0 gm of

NaOH

are contained in one

decilitre of solution. Its molarity would be (a) 4 M (b) 2 M (c) 1 M (d) 1.5 M  1 litre 4 1 Ans. (c) M = m.wt.w Volume = =1M . litre 40  0.1

138. When 90 gm of water is mixed with

300 gm of acetic acid. The total number of moles will be (a) 5 (b) 10 (c) 15 (d) 20 Ans. (b)Number of moles =

w1 w 2 90 300 + = + = 10 m 1 m 2 18 60

139. A molal solution is one that contains

one mole of a solute in (a) 1000 gm of the solvent (b) One litre of the solvent (c) One litre of the solution (d) 22.4 litres of the solution

Ans. (a)The number of moles of solute dissolved in 1000 gm of the solvent is called molal solution. 140. What

weight

of

ferrous

ammonium

sulphate is needed to prepare 100 ml of 0.1 normal solution (mol. wt. 392) (a) 39.2 gm (b) 3.92 gm (c) 1.96 gm

(d) 19.6 gm Ans.

100  392 = 3 .92 g (b) w = 0.1 1000

141. If 18 gm of glucose

(C6 H12O6 )

is present

in 1000 gm of an aqueous solution of glucose, it is said to be (a) 1 molal (b) 1.1 molal (c) 0.5 molal

(d) 0.1 molal Ans. (d) 18018 1 = 101 = 0.1 molal. 142. The number of moles of

3 molar solution is (a) 1 (b) 2 (c) 3 (d) 1.5

KCl

in

1000 ml

of

Ans.

(c)

M =

n V (l)



3=

n 1

 n

=

3 moles.

143. The unit of molality is

(a) Mole per litre (b) Mole per kilogram (c) Per mole per litre (d) Mole litre Ans. (b)The unit of molality is mole per kilogram.

144. A solution contains 1 mole of water

and 4 mole of ethanol. The mole fraction of water and ethanol will be (a) 0.2 water + 0.8 ethanol (b) 0.4 water + 0.6 ethanol (c) 0.6 water + 0.8 ethanol (d) 0.8 water + 0.2 ethanol

Ans.

(a)

0.2 water + 0.8 ethanol;

XA =

mole fraction of water,

XB =

mole fraction of ethanol

XA =

N1 N1 + N 2



,

XB =

N2 N 2 + N1

Mole fraction of water = 0.2

and ethanol = 0.8.

Colligative properties

1. The magnitude of colligative properties

in all colloidal dispersions is ….than solution

(a) Lower (b) Higher (c) Both (d) None

Ans.

(a) Lower

2. Equimolar solutions in the same solvent

have (a) Same boiling point but different freezing point (b) Same freezing point but different boiling point (c) Same boiling and same freezing points

(d) Different boiling and different freezing points Ans. (c) Same boiling and same freezing points 3. Which of the following is a colligative

property (a) Osmotic pressure (b) Boiling point

(c) Vapour pressure (d) Freezing point Ans. (a)

Osmotic pressure is colligative

property. 4. The colligative properties of a solution

depend on (a) Nature of solute particles present in it (b) Nature of solvent used

(c) Number of solute particles present in it (d) Number of moles of solvent only Ans. (c)Number of solute particles present in it 5. Which

of

the

colligative property (a) Osmotic pressure (b) Elevation in B.P.

following

is

not

a

(c) Vapour pressure (d) Depression in freezing point Ans. (c) Vapour pressure Ans. (c) Vapour pressure is not colligative property. 6. Which

of

the

colligative property (a) Optical activity

following

is

not

a

(b) Elevation in boiling point (c) Osmotic pressure (d) Lowering of vapour pressure Ans. (a)Optical activity 7. Colligative

properties

of

a

solution

depends upon (a) Nature of both solvent and solute

(b) The

relative

number

of

solute

and

solvent particles (c) Nature of solute only (d) Nature of solvent only Ans. (b)The relative number of solute and solvent particles

8.

Which is not a colligative property

(a) Refractive index

(b) Lowering of vapour pressure (c) Depression of freezing point (d) Elevation of boiling point Ans. (a)Refractive index 9. Which of the following is a colligative

property (a) Surface tension (b) Viscosity

(c) Osmotic pressure (d) Optical rotation Ans. (c) Osmotic pressure 10. Colligative properties are used for the

determination of (a) Molar Mass (b) Equivalent weight (c) Arrangement of molecules

(d) Melting point and boiling point (e) Both (a) and (b) Ans. (a)Molar Mass 11. What

does not change on changing

temperature (a) Mole fraction (b) Normality (c) Molality

(d) None of these Ans. (a)Mole fraction

Lowering of vapour pressure 1. Vapour pressure of

Hg 0.5 gm

pressure CCl 4 = 1 . 58 g / cm 2

(b) (c)

at

25 o C

is

143 mm

of

of a non-volatile solute (mol. wt. =

65) is dissolved in

(a)

CCl4

141 .43 mm

94 .39 mm

199 .34 mm

of )

the

100 ml CCl4

. Find the vapour

solution

(Density

of

(d)

143 .99 mm

Ans. (a) P P− Ps = wmWM 0

0

2. For

= 143 −

0 .5  154  143 65  158

a solution of volatile liquids the

partial vapour pressure of each component in solution is directly proportional to (a) Molarity (b) Mole fraction (c) Molality

(d) Normality Ans. (d)

P 0 − Ps P0

=

w m w W + m M

or

0 .00713 =

71 .5 m 71 .5 1000 + m 18

m = 180

3. “The

relative lowering of the vapour

pressure is equal to the mole fraction of the solute.” This law is called (a) Henry's law (b) Raoult's law

(c) Ostwald's law (d) Arrhenius's law Ans.

(b)Raoult's law

4. The relative lowering of vapour pressure

produced

by

dissolving

71.5

g

of

a

substance in 1000 g of water is 0.00713. The molecular weight of the substance will be (a) 18.0

(b) 342 (c) 60 (d) 180 Ans. (d)

P 0 − Ps P0

=

w m w W + m M

or

0 .00713 =

71 .5 m 71 .5 1000 + m 18

m = 180

5. When mercuric iodide is added to the

aqueous solution of potassium iodide, the

(a) Freezing point is raised (b) Freezing point is lowered (c) Freezing point does not change (d) Boiling point does not change Ans. (b) HgI although insoluble in water but 2

shows

complex

formation

with

freezing point is decreases. 6. Vapour pressure of a solution is

KI

and

(a) Directly proportional to the mole fraction of the solvent (b) Inversely

proportional

to

the

mole

to

the

mole

fraction of the solute (c) Inversely

proportional

fraction of the solvent (d) Directly proportional to the mole fraction of the solute

Ans.

(a)

For

solutions

containing

non-

volatile solutes, the Raoult’s law may be stated

as

at

a

given

temperature,

the

vapour pressure of a solution containing non-volatile solute is directly proportional to the mole fraction of the solvent. 7. When a substance is dissolved in a

solvent the vapour pressure of the solvent is decreased. This results in

(a) An increase in the b.p. of the solution (b) A decrease in the b.p. of the solvent (c) The solution having a higher freezing point than the solvent (d) The solution having a lower osmotic pressure than the solvent Ans.

(a)Vapour pressure 

1 Boiling po int

When

vapour

pressure

decreases

then

b.pt. increases. 8. If

Po

and

P

are the vapour pressure of a

solvent and its solution respectively fractions

of

and the

N1

and

N2

solvent

are the mole and

respectively, then correct relation is (a) (b)

P = P o N1

P = Po N2

solute

(c) (d)

Po = P N2

P = P o (N1 / N 2 )

Ans. (a) 9. An

P = P o N1

aqueous solution of methanol in

water has vapour pressure (a) Equal to that of water (b) Equal to that of methanol (c) More than that of water

(d) Less than that of water Ans.

(c)

point than

Methanol has low boiling

H 2O

Lower is boiling point of solvent more is vapour pressure. 10. The pressure under

which liquid and

vapour can coexist at equilibrium is called the (a) Limiting vapour pressure

(b) Real vapour pressure (c) Normal vapour pressure (d) Saturated vapour pressure Ans.

(b)Real vapour pressure

11. Which solution will show the maximum

vapour

pressure at 300 K

(a) 1 M

C 12 H 22O11

(b) 1 M

CH 3 COOH

(c) 1 M (d) 1 M Ans.

NaCl2

NaCl

(a)

Sucrose

minimum value of

P

will

give

.

P = P 0 − Ps

Ps = P 0 − P

is maximum.

12. The

relative

lowering

of

the

vapour

pressure is equal to the ratio between the number of

(a) Solute moleules and solvent molecules (b) Solute

molecules

and

the

total

and

the

total

molecules in the solution (c) Solvent

molecules

molecules in the solution (d) Solvent molecules and the total number of ions of the solute Ans. (b)The relative lowering of the vapour pressure of dilute solution is equal to the

mole

fraction

of

the

solute

molecule

present in the solution. 13.

5cm 3

of acetone is added to

100 cm 3

of

water, the vapour pressure of water over the solution (a) It will be equal to the vapour pressure of pure water (b) It will be less than the vapour pressure of pure water

(c) It

will

be

greater

than

the

vapour

solution

has

vapour

pressure of pure water (d) It will be very large Ans.

(b)

Acetone

pressure less than pure water. 14. At 300 K, when a solute is added to a

solvent its vapour

pressure over the mercury reduces from 50

mm to 45 mm. The value of mole fraction of solute will be (a) 0.005 (b) 0.010 (c) 0.100 (d) 0.900 Ans. (c) 0.100

15. A solution has a 1 : 4 mole ratio of

pentane to hexane. The vapour pressure of the pure hydrocarbons at 20°C are 440 mmHg for pentane and 120 mmHg for hexane. The mole fraction of pentane in the vapour phase would be (a) 0.549 (b) 0.200 (c) 0.786

(d) 0.478 Ans. (d)

=

PT = Pp0 x p + Ph0 x h

88 + 96

= 184 ;

=

440 

Pp0 x p = y p PT

1 4 + 120  5 5

;

88 = yp 184

y p = 0 .478

16. Benzene and toluene form nearly ideal

solutions. At 20°C, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The parial vapour pressure of benzene at 20°C for a solution containing

78g of benzene and 46g of toluene in torr is (a) 50 (b) 25 (c) 37.5 (d) 53.5 Ans. (a) PB = = 50 torr

PBo X B

;



PB =

78 78

78 46 + 78 92

 75

; 

PB

17. The vapour pressure lowering caused

by

the

addition

of

100

g

of

sucrose(molecular mass = 342) to 1000 g of water if the vapour pressure of pure water at

25 o C

is 23.8 mm Hg

(a) 1.25 mm Hg (b) 0.125 mm Hg (c) 1.15 mm Hg (d) 00.12 mm Hg

Ans. (b)Given molecular mass of sucrose = 342 Moles of sucrose Moles of water

N=

=

mole

100 = 0 .292 342

1000 = 55 .5 18

moles and

Vapour pressure of pure water

mm Hg According to Raoult’s law P P

0

=

n P 0 .292  = n+N 23 .8 0 .292 + 55 .5

P 0 = 23 .8

P =

23 .8  0 .292 = 0 .125 55 .792

mm Hg.

18. Which of the following is incorrect

(a) Relative lowering of vapour pressure is independent (b) The vapour pressure is a colligative property (c) Vapour pressure of a solution is lower than the vapour pressure of the solvent

(d) The relative lowering of vapour pressure is

directly

propertional

to

the

original

pressure Ans.

(d)According

to

Raoult's

law,

the

relative lowering in vapour pressure of a dilute solution is equal to mole fraction of the solute present in the solution. 19. Among

the

following

lowest vapour pressure is

substances

the

exerted by (a) Water (b) Mercury (c) Kerosene (d) Rectified spirit Ans. (b) Mercury

20. According to Raoult's law the relative

lowering of vapour pressure of a solution of volatile substance is equal to (a) Mole fraction of the solvent (b) Mole fraction of the solute (c) Weight percentage of a solute (d) Weight percentage of a solvent Ans. (b) Mole fraction of the solute

21. When a substance is dissolved in a

solvent, the vapour pressure of the solvent is decreased. This results in (a) An increase in the boiling point of the solution (b) A

decrease

in

the

boiling

point

of

solvent (c) The solution having a higher freezing point than the solvent

(d) The solution having a lower osmotic pressure than the solvent Ans. (a) When vapour pressure of solvent decreases, then the boiling point of solvent increases. 22. The

vapour

pressure

of

depends on (a) Temperature but not on volume (b) Volume but not on temperature

a

liquid

(c) Temperature and volume (d) Neither on temperature nor on volume Ans. (a)Temperature but not on volume 23. Which

one

of

the

statements

given

below concerning properties of solutions, describes a colligative effect (a) Boiling point of pure water decreases by the addition of ethanol

(b) Vapour

pressure

of

pure

water

decreases by the addition of nitric acid (c) Vapour

pressure

of

pure

benzene

decreases by the addition of naphthalene (d) Boiling point of pure benzene increases by the addition of toluene Ans. (b) Vapour pressure of pure water decreases by the addition of nitric acid 24. The atmospheric pressure is sum of the

(a) Pressure of the biomolecules (b) Vapour

pressure

of

atmospheric

constituents (c) Vapour

pressure

of

chemicals

and

vapour pressure of volatiles (d) Pressure

created

on

to

atmospheric

molecules Ans. (b) Vapour pressure of atmospheric constituents

25. The vapour pressure of pure liquid A is

0.80 atm. On mixing a non-volatile B to A, its vapour pressure becomes 0.6 atm. The mole fraction of B in the solution is (a) 0.150 (b) 0.25 (c) 0.50 (d) 0.75 Ans. (b) According to Raoult’s Law

P 0 − Ps P0

xB =

= xB

0 .8 − 0 .6 = 0.25 0 .8

(Mole fraction of solute)

.

26. Lowering of vapour pressure is highest

for (a) Urea (b)

0 .1 M

glucose

(c)

0.1 M MgSO4

(d)

0.1 M BaCl2

Ans. (d)

P 0 − Ps P0

= molality  (1 -  + x  + y  )

is maximum for 27. An

BaCl2

the value of

P 0 − Ps

.

aqueous solution of glucose was

prepared by dissolving 18 g of glucose in 90 g of water. The relative lowering in vapour pressure is (a) 0.02 (b) 1 (c) 20

(d) 180 Ans. (a)

P 0 − Ps P

0

=

18  18 = 0.02 180  90

.

28. “Relative lowering in vapour pressure of

solution

containing

directly

proportional

non-volatile to

mole

solute”. Above statement is (a) Henry law (b) Dulong and Petit law

solute

is

fraction

of

(c) Raoult's law (d) Le-Chatelier's principle Ans. (c) Raoult's law 29. An

ideal

solution

was

obtained

by

mixing methanol and ethanol. If the partial vapour pressure of methanol and ethanol are

2.619 kPa

and

4.556 kPa

respectively,

the

composition of the vapour (in terms of mole fraction) will be

(a) 0.635 methanol, 0.365 ethanol (b) 0.365 methanol, 0.635 ethanol (c) 0.574 methanol, 0.326 ethanol (d) 0.173 methanol, 0.827 ethanol Ans. (b) 0.365 methanol, 0.635 ethanol 30. The vapour pressure of two liquids P

and Q are 80 and 600 torr, respectively. The

total

vapour

pressure

of

solution

obtained by mixing 3 mole of P and 2

mole of Q would be (a) 140 torr (b) 20 torr (c) 68 torr (d) 72 torr Ans. (d)

PT = PP0 X P + PQ0 X Q

PT = 48 + 24 = 72 torr

.

;P

T

= 80 

3 2 + 60  5 5

31. The vapour pressure of benzene at a

certain temperature is

640 mm

of

Hg

. A non-

volatile and non-electrolyte solid weighing 2 . 175 g

is added to

39 .08 g

of benzene. The

vapour pressure of the solution is

Hg

600 mm

of

. What is the molecular weight of solid

substance (a) 49.50 (b) 59.6

(c) 69.5 (d) 79.8 Ans. (c)

=

m = 69 .45

P 0 − Ps P0

w M 40 2.175  78   = m W 640 m  39 .08

=

w m w W + m M

W w  M m



640 − 600 640

 78 640  ; m = 2.175 39 .08 40

.

32. Which

one

of

the

expression of Raoult's law (a)



p − ps n = p n+N

following

is

the

(b)

ps − p N = p N +n

(c)

p − ps N = ps N −n

(d)

p=

vapour pressure of pure solvent

ps =

n=

ps − p N − n = ps N

vapour pressure of the solution

number of moles of the solute

N =

number of moles of the solvent

Ans. (a) p −pp

s

=

n n+N

33. Which has maximum vapour pressure

(a) HI (b) HBr (c) HCl (d) HF Ans. (c) The lower is boiling point more is vapour

pressure;

HCl  HBr  HI  HF

boiling

point

order,

34. When a non-volatile solute is dissolved

in a solvent, the relative lowering of vapour pressure is equal to (a) Mole fraction of solute (b) Mole fraction of solvent (c) Concentration of the solute in grams per litre (d) Concentration of the solute in grams 100

ml

Ans. (a) Mole fraction of solute

35.

60

gm of Urea (Mol. wt 60) was

dissolved in 9.9 moles, of water. If the vapour pressure of pure water is vapour pressure of solution is (a) 0.10

Po

(b) 1.10

Po

(c) 0.90

Po

Po

, the

(d) 0.99

Po

Ans. (c)

8 .9 P 0 = 9 .9 Ps  Ps =

P 0 − Ps P

0

=

P 0 − Ps n 1  =  9.9 P 0 − 9.9 Ps = P 0 N 9.9 P0

8 .9 0 P  0 .90 P 0 9 .9

36. The vapour pressure of water at

17.54

20 o C

is

mm. When 20g of a non-ionic,

substance is dissolved in 100g of water, the vapour pressure is lowered by 0.30

mm. What is the molecular weight of the substances

(a) 210.2 (b)

206.88 (c) 215.2 (d) 200.8 Ans. (b) 206.88 37. In an experiment, 1 g of a non-volatile

solute was dissolved in 100 g of acetone (mol. mass = 58) at 298K. The vapour

pressure of the solution was found to be 192.5 mm Hg. The molecular weight of the solute is (vapour pressure of acetone = 195 mm Hg) (a) 25.24 (b) 35.24 (c) 45.24 (d) 55.24

Ans. (c) 45.24 38. How many grams of

CH 3 OH

added to water to prepare (a) 9.6 (b) 2.4 (c)

9 .6  10 3

(d)

2 .4  10 3

150 ml

solution of

2 M CH 3 OH

should be

1000 ml of

Ans. (a)

CH 3 OH

requires

methanol = 32 g. 150 ml of 2 M

=

32  150  2 = 9 .6 g 1000

39. The

CH 3 OH

requires methanol

.

vapour

decreased by non-volatile

pressure 10 mm

solute

of was

of

a

solvent

mercury, when a added

to

the

solvent. The mole fraction of the solute in the solution is 0.2. What should be the

mole fraction of the solvent, if decrease in the

vapour

pressure

is

mercury (a) 0.8 (b) 0.6 (c) 0.4 (d) 0.2 Ans.

(b)  P

0

− Ps = P 0  mole fraction solute

to

be

20 mm

of

10 = P 0  0 .2

;

20 = P 0  n



n = 0 .4



N = 0 .6

.

40.For a dilute solution, Raoult's law states

that (a) The lowering of vapour pressure is equal to mole fraction of solute (b) The relative lowering of vapour pressure is equal to mole fraction of solute

(c) The relative lowering of vapour pressure is proportional to the amount of solute in solution (d) The vapour pressure of the solution is equal to the mole fraction of solvent Ans. (b) According to the Raoult’s law for the non-volatile solute the relative lowering of vapour pressure of a solution containing a

non-volatile is equal to the mole

fraction of

the solute. 41. The vapour pressure of a solvent A is

0.80 atm

When a non-volatile substance

B is added to this solvent its vapour pressure drops to 0.6 atm. What is mole fraction of B in solution (a) 0.25 (b) 0.50

(c) 0.75 (d) 0.90 Ans. (a) 0.25 42. Determination of correct molecular mass

from Raoult's law is applicable to (a) An electrolyte in solution (b) A non-electrolyte in a dilute solution

(c) A

non-electrolyte

in

a

concentrated

solution (d) An electrolyte in a liquid solvent Ans.

(b)

A

non-electrolyte

in

a

dilute

solution 43. If two substances

A

and

B

have

PA0 : PB0 = 1 : 2

and have mole fraction in solution 1 : 2 then mole fraction of (a) 0.33

A

in vapours

(b) 0.25 (c) 0.52 (d) 0.2 Ans. (d) Relationship between mole fraction of a component in the vapour phase and total vapour pressure of an ideal solution.

yA =

PA x A . PA0 = Ptotal x A . PA0 + x B . PB0

=

1 1 1 1 = = = 0.2 1 1 + 2  2 1 + 4 5

44. A

dry

air

is

passed

through

the

solution, containing the 10 gm of solute and 90 gm of water and then it pass through

pure

water.

There

is

the

depression in weight of solution wt by 2.5 gm and in weight of pure solvent by 0.05 gm. solute (a) 50

Calculate

the

molecular

weight

of

(b) 180 (c) 100 (d) 25 (e) 51 Ans. (c)

Lowering in weight of

solution



solution pressure

in weight of solvent (

p0 =



Lowering

P 0 − Ps

vapour pressure of pure solvent)

p0 − ps Lowering in weight of solvent = ps Lowering in weight of solution

p0 − ps w  M = ps m W

0 .05 10  18 2  2 .5 2  250 = = = 100  m= 2 .5 90  m 0 .05 5

Ideal and Non-ideal solution 1. Which

of

the

following

liquid

pairs

shows a positive deviation from Raoult's law

(a) Water-nitric acid (b) Benzene-methanol (c) Water-hydrochloric acid (d) Acetone-chloroform Ans. (b)

In solution showing

positive type of deviation the partial pressure of each component of solution is greater than the vapour pressure as expected according to Raoult’s law.

In solution of methanol & benzene methanol molecules are held together due to hydrogen bonding as shown below. CH 3 CH 3 CH 3 | | | − − − −O — H − − − −O — H − − − −O — H − − − −

On

adding

benzene,

the

benzene

molecules get in between the molecule of methanol

thus

breaking

the

hydrogen

bonds. As the resulting solution has weak intermolecular

attraction,

the

escaping

tendency of alcohol & benzene molecule from the solution increases. Consequently the vapour pressure of the solution is greater

than

the

vapour

pressure

as

expected from Raoult’s law. 2. Which one of the following is non-ideal

solution (a) Benzene + toluene (b) n -hexane + n -heptane

(c) Ethyl bromide + ethyl iodide (d)

CCl4 + CHCl3

Ans. (d)

CCl4 + CHCl3

3. A non ideal solution was prepared by

mixing

30

ml

chloroform

and

50

acetone. The volume of mixture will be (a) > 80 ml (b) < 80 ml

ml

(c) = 80 ml (d)  80 ml Ans. (b)

Chloroform

&

acetone

form a non-ideal solution, in which

A..... B

type interaction are more than

B...... B

type interaction due to

H

A...... A

&

-bonding. Hence,

the solution shows, negative deviation from Raoult’s Law i.e.,

Vmix = −ve

;

H mix = −ve



total volume of solution = less than

(30 + 50 ml) or

 80 ml

4. Which pair from the following will not

form an ideal solution (a)

CCl4 + SiCl4

(b)

H 2 O + C4 H 9 OH

(c)

C2 H 5 Br + C2 H 5 I

(d)

C6 H14 + C7 H16

Ans. (b)

H 2O

and

C4 H 9 OH

do not form ideal

solution because there is hydrogen bonding between

H 2O

and

C4 H 9 OH

.

5. An ideal solution is that which

(a) Shows positive deviation from Raoult's law (b) Shows negative deviation from Raoult's law

(c)

Has no connection with Raoult's law

(d) Obeys Raoult's law Ans. (d) Obeys Raoult's law 6. Which one of the following mixtures can

be separated into pure components by fractional distillation (a) Benzene – toluene (b) Water – ethyl alcohol

(c) Water – nitric acid (d) Water – hydrochloric acid Ans.

(a)

Aromatic

separated

by

compound

fractional

distillation.

Benzene + Toluene. 7. All form ideal solutions except

(a)

C2 H 5 Br

and

(b)

C6 H 5 Cl

and

C2 H 5 I

C6 H 5 Br

generally

e.g.

(c)

C6 H 6

(d)

C2 H 5 I

and

C6 H 5 CH 3

and

Ans. (d)

C2 H 5 I

C2 H 5 OH

and

C2 H 5 OH

do not form ideal

solution. 8. Which property is shown by an ideal

solution (a) It follows Raoult's law (b)

H mix = 0

(c)

Vmix = 0

(d) All of these Ans. (d) All of these 9. When two liquid A and B are mixed

then their boiling points becomes greater than both of them. What is the nature of this solution (a) Ideal solution

(b) Positive deviation with non ideal solution (c) Negative

deviation

with

non

ideal

solution (d) Normal solution Ans. (c)Negative deviation with non ideal solution 10. In mixture A and B components show –

ve deviation as (a)

Vmix  0

(b)

H mix  0

(c) A-B interaction is weaker than A-A and

B-B interaction (d) A-B interaction is strong than A-A and B-B interaction Ans. (b) H 11. In

0

which

applicable (a)

mix

1M NaCl

case

Raoult's

law

is

not

(b) 1 M urea (c) 1 M glucose (d) 1 M sucrose Ans. (a)

1M NaCl

12. A solution that obeys Raoult's law is

(a) Normal (b) Molar (c) Ideal

(d) Saturated Ans. (c) Ideal 13. An example of near ideal solution is

(a) n -heptane and n -hexane (b)

CH 3 COOH + C5 H 5 N

(c)

CHCl3 + (C 2 H 5 )2 O

(d)

H 2 O + HNO3

Ans. (a) n -heptane and n -hexane

14. A

mixture of liquid showing positive

deviation in Raoult's law is (a)

(CH 3 )2 CO + C2 H 5 OH

(b)

(CH 3 )2 CO + CHCl3

(c)

(C2 H 5 )2 O + CHCl3

(d)

(CH 3 )2 CO + C6 H 5 NH 2

Ans. (a)

(CH 3 )2 CO + C2 H 5 OH

15. All form ideal solution except

(a)

C2 H 5 Br

and

C2 H 5 I

(b)

C2 H 5 Cl

and

C6 H 5 Br

(c)

C6 H 6

(d)

C2 H 5 I

and

C6 H 5 CH 3

and

Ans. (d)

C2 H 5 OH

C2 H 5 I

16. Formation

and of

C2 H 5 OH

a

solution

from

components can be considered as

two

(i) Pure

solvent

→

separated

solvent

moleculesH1 (ii)

Pure

solute

→

separated

solute

molecules H2 (iii) Separated solvent and solute molecules → solution H3 Solution so formed will be ideal if (a)

H soln = H 3 − H1 − H 2

(b)

H soln = H1 + H 2 + H 3

(c)

H soln = H1 + H 2 − H 3

(d)

H soln = H1 − H 2 − H 3

Ans. (b)

H soln = H1 + H 2 + H 3

17. Identify the mixture that shows positive

deviation from Raoult’s law (a)

CHCl3 + (CH 3 )2 CO

(b)

(CH 3 )2 CO + C6 H 5 NH 2

(c)

CHCl3 + C6 H 6

(d)

(CH 3 )2 CO + CS 2

(e)

C6 H 5 N + CH 3 COOH

Ans. (d)

(CH 3 )2 CO + CS 2

18. When acetone is added to chloroform,

then hydrogen bond is formed between them.These liquids show (a) Positive deviation from Raoult's law (b) Negative deviation from Raoult's law

(c) No deviation from Raoult's law (d) Volume is slightly increased Ans. (b) Negative deviation from Raoult's law 19. Which

of the following is true when

components forming an ideal solution are mixed (a)

H m = Vm = 0

(b)

H m  Vm

(c)

Hm  Vm

(d)

H m = Vm = 1

Ans. (a) For the ideal solution Vmix = 0

H mix

and

.

20. The liquid pair benzene-toluene shows

(a) Irregular deviation from Raoult's law

(b) Negative deviation from Raoult's law (c) Positive deviation from Raoult's law (d) Practically no deviation from Raoult's law Ans.

(d)

Practically

no

deviation

from

Raoult's law 21. The solution which shows negative or

positive deviation by Raoult's law, is called (a) Ideal solution

(b) Real solution (c) Non-ideal solution (d) Colloidal solution Ans. (c) Non-ideal solution 22. Which of the following does not show

positive deviation from Raoult’s law (a) Benzene-Chloroform (b) Benzene-Acetone

(c) Benzene-Ethanol (d) Benzene-Carbon tetrachloride Ans. (a) Benzene-Chloroform 23. Which of the following mixture shows

positive deviation by ideal behaviour (a)

CHCl3 + (CH 3 )2 CO

(b)

C6 H 6 + C6 H 5 CH 3

(c)

H 2 O + HCl

(d)

CCl4 + CHCl3

Ans. (d)

CCl4 + CHCl3

24. Which property is not found in ideal

solution (a)

PA  PAo  X A

(b)

H mix  0

(c)

Vmix  0

(d) All of these

Ans. (d) All of these 25. Which of the following is not correct for

ideal solution (a)

S mix = 0

(b)

H mix = 0

(c) It obeys Raoult's law (d)

Vmix = 0

Ans. (a) For the ideal solution

S mix

is not

equal to zero. 26. Which of the following does not show

negative deviation from Raoult’s law (a) Acetone-Chloroform (b) Acetone-Benzene (c) Chloroform-Ether (d) Chloroform-Benzene

Ans. (b) Acetone-Benzene 27. A mixture of benzene and toluene forms

(a) An ideal solution (b) Non-ideal solution (c) Suspension (d) Emulsion Ans. (a) An ideal solution

28. Which

of

the

following

is

an

solution (a) Water + ethanol (b) Chloroform + carbon tetrachloride (c) Benzene + toluene (d) Water + hydrochloric acid Ans. (c) Benzene + toluene

ideal

29. When

ethanol mixes

cyclohexane

in

cyclohexane;

reduces the intermolecular

forces between ethanol molecule. In this, liquid pair shows (a) Positive deviation by Raoult's law (b) Negative deviation by Raoult's law (c) No deviation by Raoult's law (d) Decrease in volume Ans. (a) Positive deviation by Raoult's law

30. Liquids A and B form an ideal solution

(a)

The enthalpy of mixing is zero

(b)

The entropy of mixing is zero (c) The free energy of mixing is zero (d) The free energy as well as the entropy of mixing are each zero Ans. (a) The enthalpy of mixing is zero

Azeotropic mixture 1. The azeotropic mixture of water

and

HCl (b. p. 85 o C )

boils

at

108 .5 o C

.

When

(b. p.100 o C)

this

mixture is distilled it is possible to obtain (a) Pure

HCl

(b) Pure water

(c) Pure water as well as pure (d) Neither

HCl

nor

H 2O

HCl

in their pure states

Ans. (d) Azeotropic mixture is constant boiling separate

mixture, the

it

is

not

components

possible of

to

azeotropic

mixture by boiling. 2. An azeotropic solution of two liquids

has boiling point lower than either when it

(a) Shows a negative deviation from Raoult's law (b) Shows no deviation from Raoult's law (c) Shows positive deviation from Raoult's law (d) Is saturated Ans. (c) Shows positive deviation from Raoult's law

3. A liquid mixture boils without changing

constituent is called (a) Stable structure complex (b) Binary liquid mixture (c) Zeotropic liquid mixture (d) Azeotropic liquid mixture Ans.

(d) Azeotropic mixture is a

mixture of two liquids which boils at on

particular temperature like a pure liquid and distils over in the same composition. 4. Azeotropic mixture are

(a) Constant temperature boiling mixtures (b) Those

which

boils

temperatures (c) Mixture of two solids (d) None of the above

at

different

Ans.

(a)

Constant

temperature

boiling

mixtures 5. A mixture of two completely miscible

non-ideal

liquids

which

distil

as

such

without change in its composition at a constant temperature as though it were a pure liquid. This mixture is known as (a) Binary liquid mixture (b) Azeotropic mixture

(c) Eutectic mixture (d) Ideal mixture Ans. (b) Azeotropic mixture

Osmosis and Osmotic pressure of the solution 1. If 3 gm of glucose (mol. wt. 180) is

dissolved in 60 gm of water at

15 o C

. Then

the osmotic pressure of this solution will be

(a) 0.34 atm (b) 0.65 atm (c) 6.57 atm (d) 5.57 atm Ans. (c)

 = CRT =

3  1000  0 .0821  288 = 6.56 atm 180  60

.

2. The concentration in gms per litre of a

solution of cane sugar

(M = 342 )

which is

isotonic with a solution containing urea

(M = 60 )

per litre is

(a) 3.42 (b) 34.2 (c) 5.7 (d) 19 Ans. (b) Isotonic solution =

=

w1 w2 = m 1 V1 m 2 V2

w1 6 342  6 = = = 34 . 2 342  1 60  1 60

.

6 gms

of

3. Osmotic

pressure

temperature of

300 K

is

0.0821

(a) 0.033 (b) 0.066

0 .33  10 −2

(d) 3 Ans. (c)

 = CRT

,

C=

 RT

=

at

. Find concentration in

mole/litre

(c)

atm

0 .0821 = 0 .33  10 − 2 0 .821  300

.

4. Osmotic

pressure

of

a

solution

containing 0.1 mole of solute per litre at 273 K

(a) (b)

is (in atm)

0.1  0 .08205  273 1

0.1  1  0.08205  273

(c)

1  0 .08205  273 0.1

(d)

0 .1 273  1 0 .08205

Ans. (a)

=

w 0.1  RT =  0 .0821  273 m 1

5. A solution contains non-volatile solute of

molecular mass

Mp

.Which of the following

can be used to calculate molecular mass of the solute in terms of osmotic pressure (m = Mass of solute, V = Volume of solution and

(a)

m  Mp =   VRT  

(b)

 m  RT Mp =   V 

 = Osmotic pressure)

(c)

m   Mp =    V  RT

(d)

m  Mp =   RT V

Ans. (b)

=

n m RT RT  M P =   V V 

6. The osmotic pressure of a 5% (wt/vol)

solution of cane sugar at (a) 2.45 atm (b) 5.078 atm (c) 3.4 atm

150 o C

is

(d) 4 atm Ans.

=

(b)

C=

5 1 50   1000 = 342 100 342

mol/l

50  0 .082  423 = 5 .07 atm 342

7. The

relationship

pressure at

(P2 )

and

10 g

of water is (a)

P1  P2  P3

273 K

when

sucrose

(P3 )

between 10 g

glucose

osmotic (P1 ),10 g

are dissolved in

urea 250 ml

(b)

P3  P1  P2

(c)

P2  P1  P3

(d)

P2  P3  P1

Ans. (c) thus

P2  P1  P3

P

P=

w R.T mv

1 m

.

8. In osmosis

since wvT are constant

(a) Solvent

molecules

move

from

higher

concentration to lower concentration (b) Solvent molecules move from lower to higher concentration (c) Solute molecules move from higher to lower concentration (d) Solute molecules move from lower to higher concentration

Ans. (b) In the osmosis solvent molecule move from lower concentration to higher concentration. 9. Semipermeable membrane is that which

permits the passage of (a) Solute molecules only (b) Solvent molecules only (c) Solute and solvent molecules both

(d) Neither solute nor solvent molecules Ans. (b) Solvent molecules only 10. Two solutions A and B are separated

by semi- permeable membrane. If liquid flows form A to B then (a) A is less concentrated than B (b) A is more concentrated than B (c)

Both have same concentration

(d) None of these Ans. (a)Osmosis occur from dilute solution to concentrate solution. Therefore solution A is less concentrated than B. 11. A 5% solution of canesugar (mol. wt.

=342) is isotonic with 1% solution of a substance is (a) 34.2

X

. The molecular weight of

X

(b) 171.2 (c) 68.4 (d) 136.8 Ans. (c)Molar concentration of cane sugar =

5 1000 50  = 342 100 342

Molar concentration of X =

10 50 = m 342

or

m = 68 .4

.

1 1000 10  = m 100 m

12. Which

of

properties

can

proteins

(or

the

following

provide

polymers

molar or

colligative mass

colloids)

greater precision (a) Relative lowering of vapour pressure (b) Elevation of boiling point (c) Depression in freezing point (d) Osmotic pressure

of with

(e) Rast's method Ans.

(d)

Osmotic

pressure

method

is

especially suitable for the determination of molecular masses of macromolecules such as protein & polymer because for these substances the value of other colligative properties such as elevation in boiling point or depression in freezing point are too small to be measured on the other hand

osmotic pressure of such substances are measurable. 13. The average osmotic pressure of human

blood is 7.8 bar at

37 o C

.

What is the concentration of an aqueous

NaCl

solution that could be used in the blood

stream (a) (b)

0.16 mol / L

0.32 mol / L

(c) (d)

0.60 mol / L

0.45 mol / L

Ans. (b)

 = C RT

;

C=

 RT

=

7.8 = 0  31 mol / litre  082  310

14. A solution of sucrose(molar mass = 342

g/mol) is prepared by dissolving 68.4 g of it per litre of the solution, what is its osmotic pressure (R = 0.082 lit. atm. k at 273k (a) 6.02 atm

−1

mol −1

)

(b) 4.92 atm (c) 4.04 atm (d) 5.32 atm Ans. (b)  = CRT

=

w  R  T 68 .4  0 .0821  273 = mV 342

= 4.92 atm

15. Blood has been found to be isotonic

with (a) Normal saline solution

(b) Saturated

NaCl

(c) Saturated

KCl

Ans.

solution solution

(a)Normal saline solution

(d) Saturated solution of a 1 : 1 mixture of

NaCl

and

KCl

16. If 20 g of a solute was dissolved in

500 ml of water and osmotic pressure of the solution was found to be 600 mm of

Hg at

15 o C,

then molecular weight of the

solute is (a) 1000 (b) 1200 (c) 1400 (d) 1800 Ans. (b)

=

n m / MRT RT = V V

600 20  0 .0821  288  1000 = 760 500  M

;

M = 1200

17. The

solution

osmotic is

solution of

1.66 suger

pressure atm

of 0.4% urea

and.

that

of

a

of 3.42 % is 2.46 atm.

When both the solution are mixed then the osmotic pressure of the resultant solution will be (a) 1.64 atm (b) 2.46 atm (c) 2.06 atm

(d) 0.82 atm Ans.

(c)

=

1 .66 + 2 .46 = 2 .06 atm 2

18. Blood is isotonic with

(a) 0.16 M (b) Conc.

NaCl

(c) 50 %

NaCl

(d) 30 %

NaCl

NaCl

Ans. (a) 0.16 M

19. Which

inorganic

precipitate

acts

as

semipermeable membrane or The chemical composition of semipermeable membrane is (a) Calcium sulphate (b) Barium oxalate (c) Nickel phosphate (d) Copper ferrocyanide

Ans. (d) Copper ferrocyanide ppt. acts as a semipermeable membrane. 20. The osmotic pressure of

27 o C

is

(a) 2.46 atm (b) 24.6 atm (c) 1.21 atm (d) 12.1 atm

1m

solution at

Ans. (b) Osmotic pressure = CRT

where

C = 1 m

 = CRT

= 1  0 .0821  300

= 24 .6 atm

21. Osmotic pressure of a solution can be

measured quickly

and accurately by

(a) Berkeley and Hartley's method (b) Morse's method (c) Pfeffer's method

(d) De Vries method Ans. (a) Berkeley and Hartley's method 22. The solution in which the blood cells

retain their normal form are with regard to the blood (a) Isotonic (b) Isomotic (c) Hypertonic

(d) Equinormal Ans. (a) Isotonic 23. The osmotic pressure of a solution is

given by the relation (a)

P=

RT C

(b)

P=

CT R

(c)

P=

RC T

(d)

P = RT C

Ans. (d) P = CRT or

P = RT C

24. The osmotic pressure of a solution is

directly proportional to (a) The molecular concentration of solute (b) The absolute temperature at a given concentration (c) The lowering of vapour pressure (d) All of the above

Ans. (d)  = CRT or

 P 0 − Ps  dRT  0  M  P 

 = 

25. What would happen if a thin slice of

sugar beet is placed in a concentrated solution of

NaCl

(a) Sugar beet will lose water from its cells (b) Sugar solution

beet

will

absorb

water

from

(c) Sugar beet will neither absorb nor lose water (d) Sugar beet will dissolve in solution Ans. (a) Sugar beet will lose water from its cells 26. The

osmotic

solution is given by (a)

P = Po x

pressure

of

a

dilute

(b) V = nRT (c)

P = Po N 2

(d)

P Po − P = Po Po

Ans. (b)

V = nRT

27. Which

osmotic

statement pressure

(P),

temperature (T) (a)

P

1 V

if

T

is

is constant

wrong volume

regarding (V)

and

(b)

PT

if

V

is constant

(c)

PV

if

T

is constant

(d)

PV

is constant if

Ans. (c)

PV

if

T

T

is constant

is constant

28. Isotonic solutions have

(a) Equal temperature (b) Equal osmotic pressure (c) Equal volume

(d) Equal amount of solute Ans. (b) Equal osmotic pressure 29. Which of the following associated with

isotonic solutions is not correct (a) They will have the same osmotic pressure (b) They

have

concentrations

the

same

weight

(c) Osmosis does not take place when the two

solutions

are

separated

by

a

semipermeable membrane (d) They

will

have

the

same

pressure Ans. (b) They have the same weight concentrations 30. Isotonic solution have the same

(a) Density

vapour

(b) Molar concentration (c) Normality (d) None of these Ans. (b) Molar concentration 31. A

0.6%

solution

of

urea

(molecular

weight = 60) would be isotonic with (a) (b)

0 .1 M

glucose

0.1M KCl

(c) 0.6% glucose solution (d)

0.6% KCl

solution

Ans.

(a)Isotonic solutions are those

which have same concentration. 32. The value of osmotic pressure of a 0.2

M aqueous solution at 293K is (a) 8.4 atm (b) 0.48atm

(c) 4.8 atm (d) 4.0 atm Ans. (c)

 = CRT = 0 .2  0 .0821  293 = 4 .81

33. Diffusion

of

solvent

atm. through

permeable membrane is called (a) Diffusion (b) Osmosis (c) Active absorption

a

semi

(d) Plasmolysis Ans. (b) Osmosis 34. Solutions

having

the

same

osmotic

pressure under a given set of conditions are known as (a) Hypertonic (b) Hypotonic (c) Normal

(d) Isotonic Ans. (d) Isotonic 35. At low concentrations, the statement that

equimolal solutions under a given set of experimental conditions have equal osmotic pressure is true for (a) All solutions (b) Solutions of non-electrolytes only

(c) Solutions of electrolytes only (d) None of these Ans.

(b)

Equal

osmotic

pressure

only

applicable of non-electrolytes solution at low concentration. 36. Which one of the following would lose

weight on exposure to atmosphere (a) Concentrated (b) Solid

NaOH

H 2 SO 4

(c) A saturated solution of

CO2

(d) Anhydrous sodium carbonate Ans.

(c) A saturated solution of

37. The molecular weight of

NaCl

CO2

determined

by osmotic pressure method will be (a) Same as theoritical value (b) Higher than theoritical value (c) Lower than theoritical value

(d) None of these Ans. (c) Lower than theoritical value 38. The

osmotic

pressure

of

solution

increases, if (a) Temperature is decreased (b) Solution concentration is increased (c) Number increased

of

solute

molecules

is

(d) Volume is increased Ans. (c) As soon as the solute molecules increases the osmotic pressure of solution increase. 39. At

the

same

temperature,

following

solution will be isotonic (a) 3.24 gm of sucrose per litre of water and 0.18 gm glucose per litre of water

(b) 3.42 gm of sucrose per litre and 0.18

gm glucose in 0.1 litre water (c) 3.24 gm of sucrose per litre of water and 0.585 gm of sodium chloride per litre of water (d) 3.42 gm of sucrose per litre of water and 1.17 gm of sodium chloride per litre of water

Ans. (b) 3.42 gm of sucrose per litre and 0.18 gm glucose in 0.1 litre water 40.The osmotic pressure of a decinormal

solution of

BaCl2

(a) Inversely

in water is proportional

to

its

celsius

temperature (b) Inversely proportional to its absolute temperature

(c)

Directly

proportional

to

its

celsius

temperature (d) Directly

proportional

to

its

absolute

temperature Ans. (d) Directly proportional to its absolute temperature 41. Blood cells will remain as such in

(a) Hypertonic solution

(b) Hypotonic solution (c) Isotonic solution (d) None of these Ans. (c) Living cells shrinks in hypertonic solution

(plasmolysis)

while

bursts

in

hypotonic solution (endosmosis). There is no. effect when living cells are kept in isotonic solution.

42. The

osmotic

pressure

of

a

dilute

solution is directly proportional to the (a) Diffusion rate of the solute (b) Ionic concentration (c) Elevation of B.P. (d) Flow of solvent from a concentrated to a dilute solution Ans. (b) Ionic concentration

43. The osmotic pressure in atmospheres of

10% solution of canesugar at (a) 724 (b) 824 (c) 8.21 (d) 7.21 Ans. (c)

=

w RT 10 0 .821  (273 + 69 ) =  m V 342 0 .1

V = nRT

= 8.21 atm.

69 o C

is

44. Which of the following molecules would

diffuse through a cell membrane (a) Fructose (b) Glycogen (c) Haemoglobin (d) Catalase Ans. (a) Fructose

45. Two

solutions of

and

KNO3

CH 3 COOH

are

prepared separately. Molarity

of

pressures are

both

P1

is and

0. 1 M

P2

and

osmotic

respectively. The

correct relationship between the osmotic pressures is (a)

P2  P1

(b)

P1 = P2

(c)

P1  P2

(d)

P1 P2 = P1 + P2 P1 + P2

Ans. (c)

CH 3 COOH

P1  P2

KNO3

dissociates completely while

dissociates to a small extent. Hence,

.

46. The

osmotic

pressure

of

a

dilute

solution of a non-volatile solute is (a) Directly proportional to its temperature on the centigrade scale

(b) Inversely proportional to its temperature on the Kelvin scale (c) Directly proportional to its temperature on the Kelvin scale (d) Inversely proportional to its temperature on the centigrade scale Ans.

(c)

Directly

proportional

temperature on the Kelvin scale

to

its

47. Osmotic pressure of a urea solution at

10 o C

is

500 mm .

Osmotic

pressure

of

the

solution become 105.3 mm. When it is diluted and temperature raised to extent of dilution is (a) 6 Times (b) 5 Times (c) 7 Times (d) 4 Times

25 o C.

The

Ans. (b)

V = n RT

500 V1 nR  283 = 105 .3 V2 nR  298

48. If a

0 .1 M

180) and

;

V1 1 = V2 5

so

V2 = 5V1

solution of glucose (mol. wt.

0.1 molar

solution of urea (mol. wt.

60) are placed on the two sides of a semipermeable membrane to equal heights, then it will be correct to say (a) There will be no net movement across the membrane

(b) Glucose will flow across the membrane into urea solution (c) Urea will flow across the membrane into glucose solution (d) Water will flow from urea solution into glucose solution Ans. (a) There is no net movement of the solvent

through

the

semipermeable

membrane between two solution of equal concentration. 49. At

constant temperature, the osmotic

pressure of a solution (a) Directly proportional to the concentration (b) Inversely

proportional

to

the

concentration (c) Directly proportional to the square of the concentration

(d) Directly proportional to the square root of the concentration Ans.

(a)

Directly

proportional

to

the

concentration 50. The

solution

containing

4 . 0 gm

of

a

polyvinyl chloride polymer in 1 litre of dioxane was found to have

an

osmotic

atmosphere at

300 K

,

pressure

6 .0  10 −4

the value of R used is

0.082 litre atmosphere mole

−1 −1

k

. The molecular

mass of the polymer was found to be (a)

3 .0  10 2

(b)

1.6  10 5

(c)

5 .6  10 4

(d)

6 .4  10 2

Ans. (b)

V =



4  0.0821  300 m

6  10 − 4  1 =

w RT m

;

m = 1 .64  10 5

.

51. Solvent

molecules

pass

through

semipermeable membrane is called (a) Electrolysis (b) Electrophoresis (c) Cataphoresis (d) Osmosis Ans. (d) Osmosis

the

52. If

molecular

weight

of

compound

is

increased then sensitivity is decreased in which of the following methods (a) Elevation in boiling point (b) Viscosity (c) Osmosis (d) Dialysis

Ans. (d) According to the dialysis process molecular weight increases but sensitivity decreases. 53. If solubility of

NaCl

at

20 o C

is 35

gm

per

100 gm of water. Then on adding 50 gm of

NaCl

to

the

same

volume

at

same

temperature the salt remains undissolved is (a) 15 gm (b) 20 gm

(c) 50 gm (d) 35 gm Ans. (a) 15 gm 54. Which of the following associated with

isotonic solution is not correct (a) They pressure

will

have

the

same

osmotic

(b) They

have

the

same

weight

concentration (c) Osmosis does not take place when the two

solutions

are

separated

by

a

semipermeable membrane (d) They

will

have

the

same

vapour

pressure Ans.

(b)

They

concentration

have

the

same

weight

55. If osmotic pressure of a solution is

at

273 K

, then at

546 K

2 atm

, the osmotic pressure is

(a) 0.5 atm (b) 1 atm (c) 2 atm (d) 4 atm Ans. (d) doubled.

 T

; if T is doubled



is also

56. In

osmosis

reaction,

the

volume

of

solution (a) Decreases slowly (b) Increases slowly (c) Suddenly increases (d) No change Ans. (b) Osmosis reaction are takes place in increases the volume.

57. As a result of osmosis the volume of

solution (a) Increases (c) Decreases (c) Remains constant (d) Increases or decreases Ans. (d) Increases or decreases

58. A solution of urea contain 8.6 gm/litre

(mol. wt. 60.0). It is isotonic with a 5% solution

of

a

non-volatile

solute.

molecular weight of the solute will be (a) 348.9 (b) 34.89 (c) 3489 (d) 861.2

The

Ans. (a)



For

two

solution if isotonic,

C1 = C 2

8 .6 5  1000 = 60  1 m. wt.  100



non-electrolytic

m = 348 .9

59. One mole each of urea, glucose and

sodium chloride were dissolved in one litre of water Equal osmotic pressure will be produced by solutions of (a) Glucose and sodium chloride (b) Urea and glucose

(c) Sodium chloride and urea (d) None of these Ans. (b) Both urea and glucose are nonelectrolytes but NaCl being electrolyte ionises. 60.Which

solutions

of

the

produce

following the

pressure (a) 0.1 M

NaCl solution

same

aqueous osmotic

(b) 0.1 M glucose solution (c) 0.6 g urea in 100 ml solution (d) 1.0 g of a non-electrolyte solute (X) in 50 ml solution (Molar mass of X = 200) Ans. (b) 0.1 M glucose solution 61. Which

of

the

solutions are isotonic (a)

0.01 M

glucose

following (R = 0 .082

atm

aqueous K −1 mol −1 )

(b) (c) (d)

0.01 M NaNO3

500 ml

solution containing

0.04 N HCl

Ans. (a)

0.01 M

glucose

0.3 g

urea

Elevation of boiling boint of the solvent

1. The

water is

latent

heat

9700 Cal / mole

of

vapourisation

and if the b.p. is

ebullioscopic constant of water is

of 100 o C

,

(a)

0 .513 o C

(b) 1.026 (c)

o

C

10 .26 o C

(d) 1.832

o

C

Ans. (a)

Kb =

M 1 RT 02 18  1 . 987  (373 )2 = = 0 .513 o C 1000 H V 1000  9700

2. The molal elevation constant of water

= 0 .52 o C

.

The

boiling

point

of

1.0

molal

aqueous

KCl

dissociation of (a)

KCl

), therefore, should be

100 .52 o C

(b) 101 .04

o

C

(c)

99 .48 o C

(d)

98 .96 o C

Ans. (b) 

solution (assuming complete

Tb = imkb = 0.52  1  2 = 1.04

Tb = 100 + 1 .04 = 101 . 04 o C

.

.

3. The

rise

in

the

boiling

point

of

a

solution containing 1.8 gram of glucose in 100 g

of a solvent in

0.1 o C

. The molal elevation

constant of the liquid is

(b) (c)

0.1 K / m

1K /m

(d) 10 K / m

(a)

0.01

K /m

Ans. (c)

4. If

0.15 g

Kb =

Tb 0 . 1  100 = =1K /m 1 .8 m  1000 180

.

of a solute dissolved in

15 g

of

solvent is boiled at a temperature higher by

0 .216 o C

than that of the pure solvent. The

molecular weight of the substance elevation constant for the solvent is (a) 1.01 (b) 10

(molal 2 .16 o C

) is

(c) 10.1 (d) 100 Ans. (d) m = K Tw  1000 W b

=

b

2.16  0.15  1000 0.216  15

= 100 .

5. Pressure cooker reduces cooking time

for food because (a) Heat is more evenly distributed in the cooking space

(b) Boiling

point

of

water

involved

in

cooking is increased (c) The higher pressure inside the cooker crushes the food material (d) Cooking

involves

chemical

changes

helped by a rise in temperature Ans. (b)

Due to higher pressure inside

the boiling point elevated.

6. Which

of the following statements is

correct for the boiling point of solvent containing a dissolved solid substance (a) Boiling point of the liquid is depressed (b) Boiling point of the liquid is elevated (c) There is no effect on the boiling point (d) The change depends upon the polarity of liquid

Ans. (b) Dissolution of a non-volatile solute raises the boiling pt. of a liquid. 7. When a substance is dissolved in a

solvent, the vapour pressure of solvent decreases. It brings (a) A decrease in boiling point of solution (b) An solution

increase

in

boiling

point

of

the

(c) A decrease in freezing point of the solution (d) An increase in freezing point of the solution Ans. (b) As we know that Boiling point Hence,

on



1 vapour pre ssure of liquid

decreasing

boiling point will increase.

vapour

pressure,

8. Elevation in boiling point was

6 gm

of a compound

X

0 .52 o C

was dissolved in

of water. Molecular weight of

X

is

(K b

water is 0.52 per 1000 gm of water) (a) 120 (b) 60 (c) 180 (d) 600

when

100 gm

for

Ans. (b) T

b

m=

=

100  K b  w m W

100  5 . 2  6 = 60 0 . 52  100



0 . 52 =

100  5 . 2  6 m  100

.

9. If the solution boils at a temperature

and the solvent at atemperature elevation of boiling point is given by (a)

T1 + T2

(b)

T1 − T2

(c)

T2 − T1

T2

T1

the

(d)

T1  T2

Ans. (b)

T1 − T2

10. If for a sucrose solution elevation in

boiling point is 0.1°C then what will be the boiling point of NaCl solution for same molal concentration (a) 0.1°C (b) 0.2°C

(c) 0.08°C (d) 0.01°C Ans. (b) Elevation in a boiling point is a colligative property as it depends upon the number of particles. ∆T b  n For sucrose, n = 1, ∆Tb = 0.1°C For NaCl, n = 2, ∆Tb = 0.2°C

11. The molal elevation constant is the ratio

of the elevation in B.P. to (a) Molarity (b) Molality (c) Mole fraction of solute (d) Mole fraction of solvent Ans. (b)

Tb = Kb  m

or

Kb = Tb / m

12. The

water is

molal

boiling

0 .513 o C kg mol −1

is dissolved in

point

constant

for

. When 0.1 mole of sugar

200 ml

of water, the solution

boils under a pressure of one atmosphere at (a)

100 .513 o C

(b)

100 .0513 o C

(c)

100 .256 o C

(d)

101 .025 o C

Ans. (c)

= 0 .2565 o C

,

 0.1  Tb = Kb  m = 0.513    1000  200  

Tb = 100 . 256 o C

13. Value of gas constant R is

(a) 0.082 litre atm (b)

0.987 cal (c) 8.3 J

mol −1 K −1

mol −1 K −1

(d) 83 erg

mol −1 K −1

Ans. (c) 8.3 J

14. The temperature, at which the vapour

pressure of a liquid becomes equal to the atmospheric pressure is known as (a) Freezing point (b) Boiling point (c) Absolute temperature (d) None of these Ans. (b) Boiling point

15. The

elevation

in

boiling

point

of

a

solution of 13.44g of CuCl2 in 1kg of water using

the

following

information

will

be

(Molecular weight of CuCl2 = 134.4 and Kb = 0.52 K molal−1) (a) 0.16 (b) 0.05 (c) 0.1 (d) 0.2

Ans. (a) Tb = i.Kb.m

CuCl2

⎯⎯→

Cu2+ + 2Cl−

1 (1−)

0

0 2



i = 1 + 2 Assuming 100% ionization So, i = 3 Tb = 3  0.52  0.1

= 0.156  0.16

16. When 10g of a non-volatile solute is

dissolved in 100 g of benzene, it raises boiling point by the solute is (a) 223 g (b) 233 g (c) 243 g (d) 253 g

(Kb

1o C

then molecular mass of

for benzene =2.53k-m–1)

Ans.

m=

(d)

Tb =

K b  w  1000 m W

2 .53  10  1000 K b  w  1000 = = 253 g 1  100 Tb  W

.

17. An aqueous solution containing 1g of

urea boils at

100 .25 o C

containing 3 g

. The aqueous solution

of glucose in the same

volume will boil at (Molecular weight of urea

and

respectively) (a) 100 .75

o

C

glucose

are

60

and

180

(b)

100 .5 o C

(c)

100 .25 o C

(d)

100 o C

Ans. (c) 18. When

100 .25 o C

common

salt

is

dissolved

water (a) Melting point of the solution increases (b) Boiling point of the solution increases

in

(c) Boiling point of the solution decreases (d) Both melting point and boiling point decreases Ans. (b) Common salt is non-volatile and rises the b.pt. 19. During the evaporation of liquid

(a) The temperature of the liquid will rise (b) The temperature of the liquid will fall

(c) May

rise

or

fall

depending

on

the

nature (d) The temperature remains unaffected Ans. (b) In the process of evaporation, high energy molecules leave the surface of liquid, hence average kinetic energy and consequently the temperature of liquid falls. 20. At higher altitudes the boiling point of

water lowers because

(a) Atmospheric pressure is low (b) Temperature is low (c) Atmospheric pressure is high (d) None of these Ans.

(a) The boiling occurs at lowers

temperature if atmospheric pressure is lower than 76cm Hg.

21. The elevation in boiling point for one

molal solution of a solute in a solvent is called (a) Boiling

point constant

(b) Molal elevation constant (c) Cryoscopic constant (d) None of these Ans. (b) Molal elevation constant

22. A solution of 1 molal concentration of a

solute will have maximum boiling point elevation when the solvent (a) Ethyl alcohol (b) Acetone (c) Benzene (d) Chloroform Ans. (c) Benzene

is

23. Mark

the correct relationship between

the boiling points of very dilute solutions of

BaCl2 (t1 )

and

(a)

t1 = t 2

(b)

t1  t 2

(c)

t2  t1

(d)

t2

KCl(t 2 )

, having the same molarity

is approximately equal to

t1

Ans. (b)

BaCl2

furnishes more ions than

and thus shows higher boiling point

T1  T2

KCl

.

Depression of freezing point of the solvent

1. Molal depression constant for water is

1 .86 o C

. The freezing point of a 0.05 molal

solution of a non-electrolyte in water is (a) − 1.86

o

C

(b)

− 0 .93 o C

(c) − 0.093 (d)

o

C

0 .93 o C

Ans. (c)

Tf = K f  molality = 1.86  0.05 = 0.093

°C

Thus freezing point = 0 – 0.093 = − 0 .093 o C

.

2. The

amount of urea to be dissolved in

500 ml of water (K =18.6 K

mole −1

in 100g

solvent) 0 .186 o C

to produce

a

depression of

in freezing point is

(a) 9 g (b) 6 g (c) 3 g (d) 0.3 g Ans. (c)

w = 3g

T f =

100  K  w m W

0 . 186 =

100  18 . 6  w 60  500

3. The maximum freezing point falls in

(a) Camphor (b) Naphthalene (c) Benzene (d) Water Ans. (a) Camphor has the maximum value of

K f (= 39 .7)

.

4. Which one of the following statements

is FALSE (a) The correct order of osmotic pressure for

0.01

M

compound is

aqueous

solution

BaCl2  KCl  CH3COOH 

(b) The osmotic pressure given by the equation

of

each

sucrose.

( )

of a solution is

 = MRT

where M is the

molarity of the solution.

(c) Raoult's

law

states

that

the

vapour

pressure of a component over a solution is proportional to its mole fraction. (d) Two sucrose solutions of same molality prepared in different solvents will have the same freezing point depression. Ans.

(d)

The

extent

of

depression

in

freezing point varies with the number of solute particles for a fixed solvent only and

it is a characteristics feature of the nature of

solvent

also.

So

for

two

different

solvents the extent of depression may vary even

if

number

of

solute

particles

be

dissolved. 5. Solute when dissolved in water

(a) Increases the vapour pressure of water (b) Decreases the boiling point of water (c) Decreases the freezing point of water

(d) All of the above Ans. (c) Decreases the freezing point of water 6. The

freezing

prepared from

20 gm

1.25 gm

of water is

constant is

1 . 86 K mole −1

solute will be (a) 105.7

point

of

a

solution

of a non-electrolyte and

271 .9 K

. If molar depression

, then molar mass of the

(b) 106.7 (c) 115.3 (d) 93.9 Ans. (a) Molar mass

= 105 .68 = 105 .7

=

K f  1000  w T f  W

=

1 . 86  1000  1 . 25 20  1 . 1

.

7. What is the freezing point of a solution

containing

8.1 g HBr

in

100 g

water assuming the

acid

to

be

90%

ionised

(K f

for

= 1 . 86 K mole −1 )

(a)

0 .85 o C

(b)

− 3 .53 o C

(c)

0o C

(d)

− 0 .35 o C

Ans. (b) (1HBr ⇌ − )

H + + Br −





Total = 1 + 



i = 1 +  = 1 + 0. 9 = 1. 9

water

T f = i K f  m = 1 .9  1 .86 

Tf = −3.53 o C

8. If

T f

for

Kf

8 .1 1000  = 3 .53 o C 81 100

.

value of

0 . 1m

(a) 18.6 (b) 0.186 (c) 1.86 (d) 0.0186

H 2O

is 1.86. The value of

solution of non-volatile solute is

Ans. (b)

T f = K f  m = 1.86  0.1 = 0.186

9. 1% solution of

(a)

Ca(NO3 )2

. has freezing point

0o C

(b) Less than

0o C

(c) Greater than

0o C

(d) None of the above Ans. (b) Freezing point is lowered on addition of solute in it..

10. A solution of urea (mol. mass 56g mol

–

) boils at 100.18°C

1

pressure. If

Kf

and

Kb

at the atmospheric

for water are 1.86 and

0.512K kg mol–1 respectively the above solution will freeze at (a) – 6.54°C (b) 6.54°C (c) 0.654°C (d) –0.654°C

Ans. (d)

Tb = 0.18

mK b 0 . 18 = T f m Kf

T 0 − Ts = 0 . 653

; ;T

0 .18  1 .86 = T f 0 .512

0

− Ts = 0 . 653

;

;

Tb = mK b

;

T f = 0.653

Ts = 0 − 0 . 653 o C

.

11. The molar freezing point constant for

water is

(C12 H 22O11)

1 . 86 o C mole −1

. If 342 gm of canesugar

are dissolved in 1000

the solution will freeze at (a)

− 1 .86 o C

gm

of water,

(b) 1.86

o

C

(c)

− 3 .92 o C

(d)

2 . 42 o C

Ans. (a)

 342  o T f = 1.86    = 1.86  342 

; 

Tf = −1.86 o C

.

12. An aqueous solution of a non-electrolyte

boils at

100 .52 o C

. The freezing point of the

solution will be (a)

0o C

(b)

− 1 .86 o C

(c)

1 .86 o C

(d) None of the above Ans. (b)

Tb = Kb  m

Tf = K f  m = 1.86  1 = 1.86

13. The

solution

;

i.e.

0 .52 = 0 .52  m

Tf = −1.86 o C

.

freezing point of one molal assuming

NaCl

to

be

NaCl

100%

dissociated in water is (molal depression constant = 1.86)

(a)

− 1 .86 o C

(b)

− 3.72 o C

(c)

+ 1 . 86 o C

(d)

+ 3.72 o C

Ans. (b)

For

NaCl

i=2

T f = 2 K f m = 2  1.86  1 = 3.72

Ts = T − T f = 0 − 3.72 = −3.72 C

14. Heavy water freezes at

(a)

0o C

(b)

3 .8 o C

(c)

38 o C

(d)

− 0 .38 o C

Ans. (b)

3 .8 o C

15. After adding a solute freezing point of

solution decreases to – 0.186. Calculate if

K f = 1.86

and

Kb = 0.521

.

Tb

(a) 0.521 (b) 0.0521 (c) 1.86 (d) 0.0186 Ans. (b) So

T f = K f  m

m = 0 .1

,



0 .186 = 1 .86  m

Put the value of

Tb = 0.521  (0.1) = 0.0521

m

in

Tb = Kb  m

16. Given

that

T f

is

the

depression

in

freezing point of the solvent in a solution of a non-volatile solute of molality quantity

 T f lim  m →0  m

   

is equal to

(a) Zero (b) One (c) Three (d) None of the above

m

, the

Ans. (d) None of the above 17. The freezing point of 1 percent solution

of lead nitrate in water will be (a) Below (b)

0o C

(c)

1o C

(d)

2o C

0o C

Ans. (a) Dissolution of a non-volatile solute lowers the freezing pt. of the solution

H 2 O.

18. What is the effect of the addition of

sugar on the boiling and freezing points of water (a) Both boiling point and freezing point increases

(b) Both

boiling

freezing point decreases

(c)

point Boiling

and

point

increases

and

freezing

point

decreases (d) Boiling

point

decreases

and

freezing

point increases Ans. (c)

Boiling

point

increases

and

freezing point decreases 19. During depression of freezing point in a

solution the following are in equilibrium (a) Liquid solvent, solid solvent

(b) Liquid solvent, solid solute (c) Liquid solute, solid solute (d) Liquid solute solid solvent Ans. (a) Liquid solvent, solid solvent 20. 1.00

gm of a non-electrolyte solute

dissolved in 50 gm of benzene lowered the freezing point of benzene by 0.40 K. benzene is 5.12 kg of the solute will be

Kf

for

mol–1. Molecular mass

(a)

256 g mol −1

(b)

2 .56 g mol −1

(c)

512  10 3 g mol −1

(d)

2 .56  10 4 g mol −1

Ans. (a) m=

K f  1000  w T f  WSolvent (gm )

By =

using,

5 .12  1000  1 0 .40  50

= 256 gm / mol

Hence,

molecular

mass

of

the

solute

= 256 gm mol −1

21. 0.440 g of a substance dissolved in

22.2 g of benzene lowered the freezing point of benzene by mass of the substance (a) 178.9 (b) 177.8 (c) 176.7

0.567 o C

. The molecular

(K f = 5.12 o C mol −1 )

(d) 175.6 Ans. (a)

m=

K f  w  1000 T f  W

=

5 .12  0 .440  1000 = 178 .9 0 .567  22 .2

22. Which of the following aqueous molal

solution have highest freezing point (a) Urea (b) Barium chloride (c) Potassium bromide (d) Aluminium sulphate

Ans. (a)

BaCl 2 = Ba 2 + + 2Cl − = 3

KBr = K + + Br − = 2

ions

inos

Al 2 (SO 4 )3 = 2 Al 3 + + 3 SO 42 − = 5

ions

 urea is not ionise hence it is shows highest freezing point. 23. Which will show maximum depression in

freezing point when concentration is 0.1M

(a) NaCl (b) Urea (c) Glucose (d)

K 2 SO 4

Ans. (d)

NaCl → Na+ + Cl − = 2

K 2 SO 4 → 2 K + + SO 42 − = 3

ions

ions

K2 SO 4

give maximum ion in solution so it

shows

maximum

depression

in

freezing

point. 24. The freezing point of a 0.01M aqueous

glucose solution at 1 atmosphere is

− 0 .18 o C

.

To it, an addition of equal volume of 0.002

M glucose solution will; produce a solution with freezing point of nearly (a)

− 0 .036 o C

(b)

− 0 .108 o C

(c)

− 0 .216 o C

(d)

− 0 .422 o C

Ans. (c)

T f =

K f  1000  w m W

= −0.216 o C

25. What should be the freezing point of

aqueous solution containing

1000 gm

(a)

of water (water

− 0 .69 o C

Kf

17 gm

= 1.86

of

C2 H 5 OH

deg − kg mol −1

in

(b)

− 0 .34 o C

(c)

0.0 o C

(d)

0 .34 o C

Ans. (a)

T f =

1000  1 .86  17 = 0 .69 o C 46  1000

T f = 0 − 0.69 = −0.69 o C

26. In

the

depression

of

freezing

experiment, it is found that the

point

(a) Vapour pressure of the solution is less than that of pure solvent (b) Vapour pressure of the solution is more than that of pure solvent (c) Only solute molecules solidify at the freezing point (d) Only solvent molecules solidify at the freezing point

Ans. (ad) The depression of freezing point is less than that of pure solvent and only solvent molecules solidify at the freezing point. 27. Calculate the molal depression constant

of a solvent which has freezing point and latent heat of fusion (a) 2.68 (b) 3.86

180 . 75 Jg −1

.

16 .6 o C

(c) 4.68 (d) 2.86t6 Ans. (b)

Kf =

RT f2 1000  L f

,

Tf = 273 + 16 .6 = 289 .6 K

Kf =

8 .314  289 .6  289 .6 1000  180 .75

R = 8.314 JK −1mol −1

;

L f = 180 .75 Jg −1

Colligative properties of electrolyte

1. If O.P. of 1 M of the following in water

can be measured, which one will show the maximum O.P. (a)

AgNO3

(b)

MgCl2

(c)

(NH4 )3 PO4

(d)

Na2 SO 4

Ans. (c)

(NH 4 )3 PO4

gives maximum ion. Hence,

its osmotic pressure is maximum. 2. Which of the following solution in water

possesses the lowest vapour pressure (a) (b) (c)

0.1(M) NaCl

0.1(N ) BaCl2

0.1(M) KCl

(d) None of these Ans. (b)

BaCl2

gives maximum ion hence it

is shows lowest vapour pressure. 3. Which of the following solutions in water

will have the lowest vapour pressure (a) 0.1 M, NaCl (b) 0.1 M, Sucrose (c) 0.1 M,

BaCl2

(d) 0.1 M Ans.

(d)

Na3 PO4

Na3 PO4

consist of maximum ions

hence it show lowest vapour pressure. Na 3 PO 4 → 3 Na + + PO 43 − = 4

ion.

4. The vapour pressure will be lowest for

(a) 0.1 M sugar solution (b) 0.1 M KCl solution

(c) 0.1 M Cu(NO ) solution 3 2

(d) 0.1 M

AgNO3

solution

Ans. (c) Vapour pressure of a solvent is lowered by the presence of solute in it. Lowering in vapour pressure is a colligative property i.e., it depends on the no. of particles present in the solution.

Cu(NO3 )2

give

the maximum no. of ions. (i.e., 3) so it

causes the greatest lowering in vapour pressure of water. 5. Osmotic pressure of 0.1 M solution of

NaCl

and

Na2 SO 4

will be

(a) Same (b) Osmotic pressure of more than

Na2 SO 4

than

solution will be

solution

(c) Osmotic pressure of be more

NaCl

NaCl

Na2 SO 4

solution will

(d) Osmotic pressure of than that of Ans. (c) than

NaCl

NaCl

Na2 SO 4

NaSO 4

will be less

solution

have more osmotic pressure

solution because

Na2 SO 4

gives 3

ions. 6. Which

of the following solutions has

highest osmotic pressure (a)

1 M NaCl

(b) 1 M urea

(c) 1 M sucrose (d) 1 M glucose Ans. (a)NaCl gives maximum ion hence it will show highest osmotic pressure.

7.

Which one has the highest osmotic

pressure (a) (b)

M / 10 HCl

M / 10

urea

(c) (d)

M / 10 BaCl2

M / 10

glucose

Ans. (c) 8. In

and

M / 10 BaCl2

equimolar solution of glucose,

BaCl2

NaCl

, the order of osmotic pressure is

as follow (a) Glucose

 NaCl  BaCl2

(b)

Glucose

NaCl  BaCl2 

(c)

BaCl2  NaCl 

(d) Glucose Ans. (c)

NaCl



BaCl2

Glucose

 BaCl2  NaCl



Ba 2 + + 2Cl − = 3

Na + + Cl − = 2

ion

ion

Glucose  No ionisation 

BaCl2  NaCl  Glucose

9. The osmotic pressure of which solution

is

maximum

(consider

that

solution of each 90% dissociated)

deci-molar

(a) Aluminium sulphate (b) Barium chloride (c) Sodium sulphate (d) A mixture of equal volumes of (b) and (c) Ans. (a)

Al2 (SO 4 )3

gives maximum osmotic

pressure because it is gives 5 ion. 10. At

25 o C

, the highest osmotic pressure is

exhibited by

0 .1 M

solution of

(a) (b)

CaCl2

KCl

(c) Glucose (d) Urea Ans. (a) Highest osmotic pressure is given by

solution

which

number of ions i.e. 11. Which of

CaCl2

produce

maximum

.

the following will have the

highest boiling point at 1 atm pressure

(a) (b) (c) (d)

0.1 M NaCl

0 .1 M

sucrose

0.1M BaCl2

0 .1 M

glucose

Ans. (c) BaCl gives maximum ion. Hence, its 2

shows highest boiling point. 12. Which

one

of

the

following

would

produce maximum elevation in boiling point

(a) 0.1 M glucose (b) 0.2 M sucrose (c) 0.1 M barium chloride (d) 0.1 M magnesium sulphate Ans. (c)

BaCl2

gives maximum ion. Hence,

its boiling point is maximum. 13. Which

of the following solutions will

have the highest boiling point

(a) 1% glucose (b) 1% sucrose (c) 1% (d) 1%

NaCl

CaCl2

Ans. (d)

CaCl2

gives maximum ion hence it

shows highest boiling point. 14. Which

one of the following aqueous

solutions will exhibit highest boiling point

(a) 0.015 M urea (b)

0.01 M KNO3

(c)

0.01 M Na2 SO 4

(d)

0.015 M

glucose

Ans. (c) Elevation in boiling point is a colligative property which depends upon the number of solute particles. Greater the number of solute particle in a solution

higher the extent of elevation in boiling point. Na SO 2

4

15. Which

→ 2 Na + + SO 42 −

of

the

following

aqueous

solutions containing 10 gm of solute in each case has highest B.P. (a)

NaCl

(b)

KCl

solution solution

(c) Sugar solution (d) Glucose solution

Ans. (a) NaCl

contain highest boiling point

than other’s compound. 16. 0.01 molar solutions of glucose, phenol

and potassium chloride were prepared in water. The boiling points of (a) Glucose solution = Phenol solution = Potassium chloride solution (b) Potassium chloride solution > Glucose solution > Phenol solution

(c) Phenol solution > Potassium chloride solution > Glucose solution (d) Potassium chloride solution > Phenol solution > Glucose solution Ans. (d)

KCl  C 6 H 5 OH  C 6 H 12 O 6 Boiling po int decreasing order →

Potassium chloride is ionic compound and phenol is formed phenoxide ion hence it is shows greater boiling point then glucose.

17. Which one has the highest boiling point

(a)

0.1 N Na2 SO 4

(b)

0.1 N MgSO4

(c)

0.1M Al2 (SO 4 )3

(d)

0.1M BaSO4

Ans. (c)

Al2 (SO 4 )3

gives maximum ion hence it

will show highest boiling point.

18. Which of the following solutions boils at

the highest temperature (a) 0.1 M glucose (b) 0.1 M NaCl (c) 0.1 M

BaCl2

(d) 0.1 M Urea Ans. (b) NaCl is a more ionic compare to

BaCl2

, glucose and urea solution.

19.

0 .01 M

and

solution each of urea, common salt

Na2 SO 4

are taken, the ratio of depression

of freezing point is (a) 1 : 1 : 1 (b) 1 : 2 : 1 (c) 1 : 2 : 3 (d) 2 : 2 : 3 Ans. (c)Urea = 1 ; Common salt = 1 ;

Na2 SO 4 = 3

Ratio = 1 : 2 : 3 20. Which has the minimum freezing point

(a) One molal

NaCl

(b) One molal

KCl

(c) One molal

CaCl2

solution solution solution

(d) One molal urea solution

Ans. (c) CaCl

2

gives maximum ion hence it

has minimum freezing point. 21. Which

of

the

following

has

freezing point (a) (b) (c) (d)

0. 1 M

0. 1 M

0.1M

0. 1 M

aqueous solution of glucose aqueous solution of aqueous solution of

NaCl

ZnSO4

aqueous solution of urea

lowest

Ans. (b) NaCl gives maximum ion hence it shows lowest freezing point 22. The

freezing

points

solutions of glucose, the order of (a)

AlCl3  KNO3  Glucose

(b)

Glucose  KNO3  AlCl3

(c)

Glucose  AlCl3  KNO3

KNO3

of and

equimolar

AlCl3

are in

(d)

AlCl3  Glucose  KNO3

Ans. (a)

AlCl3  KNO3  Glucose

23. Which of

the following will have the

highest F.P. at one atmosphere (a) (b)

0.1M NaCl

0 .1 M

solution

sugar solution

(c)

0.1M BaCl2

solution

(d)

0.1M FeCl3

solution

Ans. (b) Lesser the number of particles in solution.

Lesser the depression in freezing

point, i.e. higher the freezing point. 24. Which of the following will produce the

maximum depression in freezing point of its aqueous solution (a)

0 .1 M

(b)

0 .1 M

glucose sodium chloride

(c)

0 .1 M

barium chloride

(d)

0 .1 M

magnesium sulphate

Ans. (c) shows

BaCl2

gives maximum ion hence it

maximum

depression

in

freezing

point. 25. Which of the following has the lowest

freezing point (a) 0.1 m sucrose

(b) 0.1 m urea (c) 0.1 m ethanol (d) 0.1 m glucose Ans. (c) 0.1 m ethanol 26. Which

of the following has minimum

freezing point (a)

0 .1 M K2 Cr2 O7

(b) 0.1 M

NH 4 Cl

(c) 0.1 M

BaSO4

(d) 0.1 M

Al 2 (SO 4 )3

Ans. (d) We know that lowering of freezing point is a colligative property which is directly

proportional

particles

formed

to by

the

number

of

one

mole

of

compound therefore 0.1M

Al2 (SO 4 )3

will have minimum freezing point.

solution

27. Which

of the following

0.10 m

aqueous

solution will have the lowest freezing point

(a)

Al 2 (SO 4 )3

(b)

C5 H10 O5

(c) (d)

KI

C12 H 22O11

Ans. (a)

Al2 (SO 4 )3

gives maximum ion hence

its gives lowest freezing point. 28. For

0.1 M

solution, the colligative

property will follow the order (a)

NaCl  Na2 SO 4  Na3 PO4

(b)

NaCl  Na2 SO 4  Na3 PO4

(c)

NaCl  Na2 SO 4  Na3 PO4

(d)

NaCl  Na2 SO 4 = Na3 PO4

Ans.

(b)Colligative property in decreasing

order

Na3 PO4  Na2 SO 4  NaCl

Na 3 PO 4 → 3 Na + + PO 43 − = 4

Na 2 SO 4 → 2 Na + + SO 42 − = 3

NaCl → Na+ + Cl − = 2

29. Which of

the following will have the

lowest vapour pressure

(a) (b)

0.1M KCl

0 .1 M

solution

urea solution

(c)

0.1M Na2 SO 4

(d)

0.1M K 4 Fe(CN )6

Ans. (d)

solution solution

K4 [Fe(CN )6 ]

gives maximum ion. Hence

it have lowest vapour pressure.

Abnormal molecular mass 1. The Van't Hoff factor will be highest for

(a) Sodium chloride (b) Magnesium chloride (c) Sodium phosphate (d) Urea

Ans.

(c)

Na3 PO4

gives maximum four ion it

is show highest Vant’s haff factor. 2. Which

of the following salt has the

same value of Van't Hoff factor of (a) (b) (c)

K 3 [Fe(CN )6 ]

Al 2 (SO 4 )3

NaCl

Na2 SO 4

i

as that

(d)

Al(NO3 )3

Ans.

(a)

thus

1

K4 [Fe(CN )6 ]

dissociates

molecule

as

dissociates

particles in the similar way

Al2 (SO 4 )3

4 K + + [Fe (CN )6 ]4 − ,

into

five

also gives

five particles per molecule. 3. When benzoic acid dissolve in benzene,

the observed molecular mass is (a) 244 (b) 61

(c) 366 (d) 122 Ans.

(a)

Benzoic

undergoes

acid

in

association

benzene through

intermolecular hydrogen bonding. 4. The ratio of the value of any colligative

property for

KCl

solution is nearly (a) 1

solution to that for sugar

(b) 0.5 (c) 2.0 (d) 3 Ans. (c)vont’s Hoff factor

= 1 −  + x + y,

(i )

=

experiment al C.P. Calculated C.P.

for KCl it is = 2 and for sugar

it is equal to 1. 5. Van't Hoff factor of

(a) 1

Ca(NO3 )2

is

(b) 2 (c) 3 (d) 4 Ans. (c)

Ca ( NO 3 )2 → Ca 2 + + 2 NO 3−

it gives three ions

hence the Van’t Hoff factor = 3. 6. Dry

air

was

through a solution of

passed 5 gm

successively

of a solute in

80 gm

of water and then through pure water. The loss in weight of solution was

2.50 gm

and

that of pure solvent

0.04 gm

. What is the

molecular weight of the solute (a) 70.31 (b) 7.143 (c) 714.3 (d) 80 Ans. (a)

m=

5  18  2 .5 = 70 .31 0 .04  80

7. The Van’t Hoff factor calculated from

association data is always...than calculated from dissociation data (a) Less (b) More (c) Same (d) More or less Ans. (a) Less

8. If

Na2 SO 4



is the degree of dissociation of

, the Vant Hoff's factor (i) used for

calculating the molecular mass is (a)

1 +

(b) (c)

1 + 2

(d)

1 − 2

Ans. (c) Na SO ⇌ 2 Na 2

4

+

+

SO 42 −

Mol. before diss. 1 Mol. after diss

1 −

0 0

2

1

i=

Exp.C.P. = 1 −  + 2 +  = 1 + 2 Normal C.P.

9. Van't Hoff factor

(a)

=

Normal molecular mass Observed molecular mass

(b)

=

Observed molecular mass Normal molecular mass

i

(c) Less than one in case of dissociation

(d) More than one in case of association Ans. (a) 10. Which

=

Normal molecular mass Observed molecular mass

of

the

following

corresponds Van't Hoff factor to 2 for dilute solution (a)

K2 SO 4

(b)

NaHSO4

(c) Sugar

compounds ' i'

to be equal

(d)

MgSO4

Ans. (d)

MgSO4

dissociates to give 2 ions.

11. The Van't Hoff factor

i

aqueous solution of urea is (a) 0.2 (b) 0.1 (c) 1.2 (d) 1.0

for a 0.2 molal

Ans. (d) Urea does not give ion in the solution. 12. One mole of a solute

a

given

volume

association according to

i

of

nA

the

i=1− x

(b)

i=1+

x n

is dissolved in

a

solvent.

solute

take

The place

⇄ (A) . The Van't Hoff factor

is expressed as

(a)

of

A

n

(c)

i=

(d)

i=1

1− x +

x n

1

Ans. (c)

i=

1− x +

x n

1

13. Acetic acid dissolved in benzene shows

a molecular weight of (a) 60 (b) 120 (c) 180

(d) 240 Ans. (b)Molecular weight of

CH 3 COOH = 60

Hence the molecular weight of acetic acid in benzene = 2  60 = 120 . 14. The

observed osmotic pressure of a

solution of benzoic acid in benzene is less than its expected value because (a) Benzene is a non-polar solvent

(b) Benzoic acid molecules are associated in benzene (c) Benzoic acid molecules are dissociated in benzene (d) Benzoic acid is an organic compound Ans.

(b)

Benzoic

acid

molecules

are

associated in benzene 15. The

experimental molecular weight of

an electrolyte will always be less than its

calculated value because the value of Van't Hoff factor “i” is (a) Less than 1 (b) Greater than 1 (c) Equivalent to one (d) Zero Ans. (b) Greater than 1

16. The

molecular

mass

dissolved

in

water

is

dissolved

in

benzene

difference in behaviour of

(a) Water

prevents

of

acetic

acid

60

and

when

it

is

CH 3 COOH

association

120.This is because

of

acetic

acid (b) Acetic acid does not fully dissolve in water

(c) Acetic acid fully dissolves in benzene (d) Acetic acid does not ionize in benzene Ans. (b) Acetic acid does not fully dissolve in water 17. The

correct

relationship

between

the

boiling points of very dilute solutions of

AlCl3 (t1 )

and

CaCl2 (t 2 )

concentration is (a)

t1 = t 2

, having the same molar

(b)

t1  t 2

(c)

t 2  t1

(d)

t 2  t1

Ans. (b) and thus t1  t 2

AlCl3

furnishes more ions than

CaCl2

shows higher boiling point i.e.

.

18. The

Van't

Hoff

phosphate would be

factor

for

sodium

(a) 1 (b) 2 (c) 3 (d) 4 Ans. (d)

Na 3 PO 4 = 3 Na + + PO 33 −

.

19. The molecular weight of benzoic acid in

benzene as determined by depression in freezing point method corresponds to

(a) Ionization of benzoic acid (b) Dimerization of benzoic acid (c) Trimerization of benzoic acid (d) Solvation of benzoic acid Ans. (b) Benzoic acid dimerises due to strong hydrogen bonding.