Solutions

Marks Allotted 13 Marks

Part-A-I 1Q (MCQ) 1M
Part-A-II 1Q (Fill in the blank) 1M
Part-B-III 1Q (Short Answer) 2M
Part-C-V 1Q (Short Answer) 3M
Part-E-VII 2Q (Numerical Problem) 6M
Total 13 M

A solution is a homogeneous mixture of 2 or more substances.

Solute + Solvent = Solution

SOLUTE – the component which is present in small quantity.

SOLVENT – the component which is present in large quantity.

Types of solutions

Types of Solute Solvent Example

solutions
Solid Solid Solid 22 Carat gold , Cu in Gold
Solutions Liquid Solid Gold Amalgam- Hg in Gold
Gas Solid Solution of Hydrogen in Pd
Liquid Solid Liquid Salt or Sugar in water
Solutions Liquid Liquid Ethanol in water
Gas Liquid O2 or CO2 in water
Gaseous Solid Gas Camphor in Oxygen
Solutions Liquid Gas Moist Hydrogen
Gas Gas Mixture of Nitrogen and Oxygen

Methods of expressing the concentration of solutions

Molarity is defined as the number of moles of solute dissolved in one dm3 or one litre
of the solution.

Mass of solute per dm3 of solution

Molarity = Molecuar mass of solute

Molality is the defined as the number of mole of solute dissolved in one Kg of the
solvent.

Mass of solute per Kg of Solvent

Molality = Molecuar mass of solute

Mass percentage is defined as the mass of solute dissolved in 100 grams of the
solution.
mass of solute
Mass percentage = mass of solution x 100

1
Volume percentage defined as the volume in cm3 of solute dissolved in 100 cm3 of the
solution.
Volume of solute
Volume percentage = x 100
Volume of solution
Mole fraction: Mole fraction of a component in a mixture is the ratio of number of
moles one component to the total number of moles of all the components present in
the solution. OR

Number moles of one component

Mole fraction = Total number of moles of all the component

Parts per million (ppm) is defined as the number of parts of mass the solute per
million parts by mass the solution. OR
Mass of the solute
PPM = Mass of the solution x 106

Solubility
Amount of solute that be dissolved in a 100 gram of solvent at a given temperature.

Saturated solution – Which contain exact amount of solute

Unsaturated solution – Which contain lesser amount of solute
Super-saturated solution – Which contain more amount of solute

Solubility of Solid in Liquid

Factors Affecting the Solubility

1. Nature of Solute and Solvent
2. Temperature

Effect of Nature of Solute and Solvent on Solubility

Polar Solutes Dissolves in Polar Solvents. (Like Dissolves Like)

Example: NaCl, KNO3, Sugar dissolves in Water.

But, Naphthalene and Anthracene does not dissolve in Water. Because, Naphthalene
and Anthracene are non-polar Solutes.

Non-Polar Solutes Dissolves in Non-Polar Solvent (Unlike Dissolves in Unlike)

Naphthalene and Anthracene dissolve in Benzene. Naphthalene, Anthracene and
Benzene are Non-Polar

Effect of Temperature on Solubility

1. If dissolution is endothermic, i.e., ∆H > 0 or ∆H is positive, then increase in
temperature increases the solubility.
Example: NaNO3, NH4Cl, KCl, AgNO3 .
2. If dissolution is exothermic, i.e., ∆H < 0 or ∆H is negative, then increase in
temperature decreases the solubility.
Example: Li2SO4, Ce2(SO4)3, Na2CO3 . H2O and Calcium Salts of some organic acid.

Solubility of Gases in Liquid

1. Effect of temperature

2
Solubility of gases decreases with increase of temperature. Because, dissolution of gas
in liquid is exothermic process.

2. Effect of Pressure
Increase in pressure increases the solubility of gases in liquids.

Henry’s law
Henry’s law states that partial pressure of the gas in vapour phase is directly
proportional to mole fraction of the gas in the solution.
p = KH x χ
Where p is the partial pressure of the gas over the solution
χ is the mole fraction and K H is is Henry’s constant.

Significance of Henry’s Constant

Higher the value of KH , lower is the solubility.
The value of KH , increases with increase in temperature. i.e., solubility of gas
decreases with increase in temperature.

Applications of Henry’s Law

To increase the solubility of carbon dioxide in soda water, soft drinks, the containers
sealed under high pressure.
At high altitudes, the partial pressure of oxygen is low. As a result, concentration of
oxygen becomes low in blood. Hence people become weak and the thinking ability
reduces. These symptoms are called anoxia.

Vapour Pressure of Liquid Solutions

Raoult’s law for a solution of volatile liquids
For a solution of volatile liquids, the partial vapour pressure of each component in the
solution is directly proportional to its mole fraction.

Graph of Vapour pressure against mole fraction of two volatile liquids

o
pB

o
pA
pressure
Vapour
pressur
Vapour

χA = 1 χA = 0
Mole Fraction
χB = 0 χB = 1

Raoult’s Law as a special case of Henry’s Law

According to Henry’s Law, p = KH X

According to Raoult’s Law, pA = pAo χA

If, KH = pAo , Raoult’s law becomes a special case of Henry’s law.

Ideal and Non-ideal Solutions

3
Ideal Solutions
Intermolecular force of attraction between the components are similar to those in the
pure components.

Non-ideal Solutions
Intermolecular force of attraction between the components are different from those in
the pure components.

Differences between ideal and non-ideal solutions

Ideal solution Non-ideal solution

Obeys Raoult’s law Does not obeys Raoult’s law
∆ Hmix = 0 ∆ Hmix ‡ 0
∆ Vmix = 0 ∆ Vmix ‡ 0
n-Hexane + n-Heptane, Ethyl alcohol + Water
Bromoethane + Chloroethane, Acetone + Carbon disulphide
Benzene + Toluene Acetone + Benzene
Chlorobenzene + Bromobenzene Acetone + Ethyl alcohol

Differences between Non-ideal Solutions showing positive and negative

deviations

Non-Ideal solution showing Non-Ideal solution showing

positive deviation negative deviation
P > pAo χA + pBo χB P < pAo χA + pBo χB
∆ Hmix > 0 i.e. + ve ∆ Hmix < 0 i.e. - ve
∆ Vmix > 0 i.e. + ve ∆ Vmix < 0 i.e. - ve
Acetone + Ethyl alcohol Chloroform + Diethyl ether
Acetone + Benzene HCl + water, HNO3 + Water
Ethyl alcohol + Water

Graph of Vapour pressure against mole fraction of Solutions showing positive

deviation

o
pB
Vapour pressure
Vapour pressure

o
pA

χA = 1 Mole Fraction χA = 0
χB = 0 χB = 1

Graph of Vapour pressure against mole fraction of Solutions showing negative

deviation

4
 o
p
B
Vapour pressure

Vapour pressure
o
p
A

χ =1 Mole χ =0
A A
χ =0 Fraction χ =1
B B

Azeotropes (A = No Zeo = Boil tropes = Change)

Solutions having the same composition in liquid and vapour phase and boil at a
constant temperature.

1. Minimum boiling azeotrope. The solutions which show a large positive deviation
from Raoult’s law. Ex. Ethanol-Water

2. Maximum boiling azeotrope- The solutions which show a large negative deviation
from Raoult’s law. Example- 68% HNO3 and 32% water by mass

Colligative properties
Properties whose values depend only on the number of solute particles present in the
solution and not on its nature, size or chemical composition.
Examples.
1. Relative lowering of vapour pressure
2. Depression of freezing point
3. Elevation of boiling point
4. Osmotic pressure

Raoult’s Law of relative lowering of vapour pressure

For a binary solution containing non-volatile solute in a volatile solvent, the relative
lowering of vapour pressure is equal to the mole fraction of the solute present in the
solution.

p0A − p
= χB
p0A
Relation between molar mass of solute and relative lowering vapour pressure.
Let 𝑀𝐴 is the molar mass of solvent.
Let 𝑊𝐴 is the mass of solvent.
Let 𝑀𝐵 is the molar mass of non-volatile solute.
Let 𝑊𝐵 is the mass of non-volatile solute.
n𝐁
Mole fraction of the solute χB = n +n
A B

Where nA & nB are the number of moles of solvent and solute respectively.
For dilute solutions, nA + nB  nA

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 n𝐁
χB =
nA

WB
MB WB MA
χB = χB =
WA MB WA
MA
But,

p0A − p
= χB
p0A
p0 − p WB MA
 A 0 =
pA MB WA

WB MA
MB =
p0 − p
WA ( A 0 )
pA

Elevation of boiling point

Solvent
Solution

Vapour
pressu
re

Tbo Δ𝑇𝑏
𝑇𝑏

Temperature
Let Tbobe the boiling point of the solvent,
Let Tb be the boiling point of the solution,
Elevation in the boiling point is given by Δ𝑇𝑏 = Tb − Tbo
But , Δ𝑇𝑏 ∝ 𝑚 Where m is molality
Δ𝑇𝑏 = K b 𝑚 where is 𝐾𝑏 called boiling point elevation constant or molal elevation
constant or ebullioscopic constant.
Let 𝑊𝐴 is the mass of solvent in kg.
Let 𝑀𝐵 is the molar mass of non-volatile solute.
Let 𝑊𝐵 is the mass of non-volatile solute.
Molality of solution is given
1000 WB
m =
WA MB
Δ𝑇𝑏 = K b 𝑚
Kb 1000 WB 1000 Kb WB
Δ𝑇𝑏 = MB =
WA MB WA Δ𝑇𝑏

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Molal elevation constant: Molal elevation constant is defined as the elevation in
boiling point produced when one mole of the solute is dissolved in 1 kg of the solvent.
Unit for molal elevation constant Kb = K kg mole-1

Depression of freezing point

Liquid
Solvent

Frozen
Solvent Solution
Vapour
Pressur
e

Tf Tfo

Temperature

Let Tfo be the freezing point of the solvent,

Let Tf be the freezing point of the solution,
Depression in the freezing point is given by ΔTf = Tf0 − Tf
But, ΔTf ∝ m Where m is molality
ΔTf = K f m Where is 𝐾𝑓 called freezing point depression constant or molal depression
constant or cryoscopic constant.
Let 𝑊𝐴 is the mass of solvent in kg.
Let 𝑀𝐵 is the molar mass of non-volatile solute.
Let 𝑊𝐵 is the mass of non-volatile solute.
Molality of solution is given
1000 WB
m =
WA MB
ΔTf = K f m
Kf 1000 WB 1000 Kf WB
ΔTf = MB =
AW BM W ΔT A f

Molal depression constant: Molal depression constant is defined as the depression in

freezing point produced when one mole of the solute is dissolved in 1 kg of the solvent.
Unit for molal depression constant Kf = K kg mole-1

Osmosis
The phenomenon of passage of solvent molecules from solvent to the solution or
movement of solvent molecules from a dilute solution to the concentrated solution
through semi permeable membrane is called osmosis.

Osmotic pressure
The external pressure which can be applied to solution in order to stop the flow of
solvent into the solution through semi permeable membrane is called osmotic
pressure.

Relationship between molar mass of a solute and osmotic pressure

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Osmotic pressure is denoted by 𝜋.
Osmotic pressure is given by 𝜋𝑉 =𝑛𝑅𝑇

n
π= RT
V

If 𝑊𝐵 is amount of solute and 𝑀𝐵 is the molar mass of mass dissolved in V liters of the
solution,

𝑊𝐵
𝑛=
𝑀𝐵

RT 𝑊𝐵 RT RT
π=n π= 𝑀𝐵 = 𝑊𝐵
V 𝑀𝐵 V πV

Isotonic solutions
The solutions having same osmotic pressure at a given temperature is called isotonic
solutions.

Hypertonic solutions
A solution which has higher osmotic pressure than the other is called hypertonic
solution.

Hypotonic solutions
A solution which has lower osmotic pressure than the other is called hypotonic
solutions.

Note:
1. 0.9% (mass/volume) sodium chloride solution is called normal saline solution
2. Blood is isotonic with 0.9% (mass/volume) sodium chloride solution
3. When blood cells are placed in a solution containing more than 0.9% sodium
chloride, water will flow out of the cells and they would shrink.
4. When blood cells are placed in a solution containing less than 0.9% sodium
chloride, water will flow in to the cells and they would swell.

Reverse osmosis
When a pressure which is greater than osmotic pressure is applied on the solution,
the solvent molecules flow from the solution into the solvent through semi permeable
membrane. This process is called reverse osmosis.

Abnormalities in molecular mass

1. Due to dissociation of solute
For NaCl,
Observed molecular mass = half of theoretical value

NaCl Na+ + Cl-

Observed colligative property = Twice of theoretical value

1
Colligative Property =
Molar mass
8
2. Due to Association of solute
For C6H5COOH,
observed molecular mass = twice the theoretical value

C6H5COOH ½ ( C6H5COOH )2

Observed colligative property = half of theoretical value

1
Colligative Property =
Molar mass

Van’t Hoff factor

normal molar mass

𝑖= or
observed molar mass

observed colligative property

𝑖= or
normal colligavibe property

total number of particles after dissociation/association

i=
number of particles before dissociation/association

For solutes showing,

1. Dissociation, i > 1
2. Association, i < 1
3. Neither dissociation nor association, i = 1

For solutes which undergoes association or dissociation,

p0A − p
1. = i χB
p0A
2. ∆Tb = i Kb m

3. ∆Tf = i Kf m

4. π = i CRT

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 Solution

Questions carrying one mark each

1. A binary liquid mixture that forms maximum boiling azeotrope at a specific

composition is (March 2023)
a. Ethanol + Water b) n - Hexane + n- Heptane
c) Benzene + Toluene d) Nitric acid + Water
Ans. a. Ethanol + Water or d. Nitric acid + Water

2. The value of van’t Hoff factor (i) for ethanoic acid in benzene is nearly (March
2023)
a. 2 b. 1 c. 0.5 d. 0
Ans. Option c) OR 0.5 OR 1

3. A beaker contains a solution of substance ‘A’. Precipitation of substance

‘A’takes place when small amount of ‘A’ is added to the solution. The solution
is
a. saturated (b) supersaturated (c) unsaturated (d) concentrated

4. The temperature independent unit of concentration is

a. weight volume percentage (b) molarity (c) normality (d) molality

5. What quantity of NaCl is to be added to 100 g of water in order to prepare

0.1 molal sodium chloride solution?
(a) 5.85 g (b) 0.585 g (c) 0.295 g (d) 2.95 g
Mass of solute x 1000
Molality =
Molar mass of solute x Mass of the solvent in Kg

0.1 x 58.5 x 100

Mass of solute = = 0.585
1000

6. Value of Henry’s constant KH.

a. increases with increase in temperature.
b. decreases with increase in temperature.
c. remains constant.
d.first increases then decreases.

7. KH value for Ar(g), CO2(g), HCHO(g) and CH4(g) are 40.39, 1.67, 1.83 x 10–5
and 0.413 respectively. Arrange these gases in the order of their increasing
solubility is.
(a) HCHO < CH4< CO2<Ar (b) HCHO < CO2< CH4<Ar
(c) Ar < CO2< CH4< HCHO (d) Ar< CH4< CO2< HCHO

8. Which one of the following conditions is not true for an ideal solution?
a. Raoult’s law (b) ∆Hmix= 0 (c) ∆Vmix = 0 (d) ∆Smix = 0

9. According to Raoult’s law, relative lowering of vapour pressure of a solution of

10
 non- volatile solute is
a. equal to mole fraction of solvent(b) directly proportional to mole fraction of solute
(c) equal to molar mass of solute (d) equal to mole fraction of solute

10. 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

11. Azeotropic mixtures are

a. mixture of two liquids with almost same boiling points
b. mixture of two liquids having different boiling points
(c)constant boiling mixtures
(d) ideal solutions in all respects

12. Which of the following is independent of temperature?

a. lowering of vapour pressure (b) vapour pressure
(c) relative lowering of vapour pressure (d) osmotic pressure

13. The relative lowering of vapour pressure caused by dissolving 1 mole of

glucose in 99 mole of water is
(a) 0.99 (b) 0.01 (c) 0.1(d) 0.06

14. The mass of glucose that should be dissolved in 50 g of water in order to

produce the same lowering of vapour pressure is produced by dissolving 1 g of
urea in the same quantity of water is
(a) 3 g (b) 1 g (c) 18 g (d) 6 g

15. The vapour pressure of a solution of 5 g of non-electrolyte in 100 g of water at

a particular temperature is 2985 Nm–2. The vapour pressure of pure water
at that temperature is 3000 Nm–2. The molecular weight of the solute is
(a) 180 (b) 90 (c) 270 (d) 200

16. The ebullioscopic constant is the elevation in boiling point produced by

a. 1 Molar solution (b) 1 Molal solution
(c) 1N solution (d) 10% solution

17. At higher altitudes the boiling point of water decreases because

a. the atmosphere pressure is high (b) the temperature is low
(c)the atmosphere pressure is low (d) the temperature is high

18. Which solution has highest boiling point?

a. 0.1 M NaCl (b) 0.1M FeCl3 (c) 0.2 M urea (d) 0.1MNa2SO4
19. The boiling points of liquids A, B, C and D are 80°C, 65°C, 184°C and 212°C
respectively. The order of their vapour pressure at room temperature is
a. D > C > A > B (b) B > A >C > D (c) A > B > C > D (d) D > C > B > A

20. Which of the following 0.1 M aqueous solutions will have the lowest freezing
point?
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 a. K2SO4 (b) NaCl (c) urea (d) glucose

21. Ethylene glycol is used an anti-freeze, since

a. it lowers the freezing point of water(b) it elevates the boiling point of water
(c) improves fuel efficiency (d) to prevent corrosion

22. Osmotic pressure of the solution can be increased by

a. increasing temperature of the solution
(b) decreasing temperature of the solution
(c) increasing the volume of the vessel
(d) diluting the solution

23. Which one of the following pair of solutions are isotonic?

a. 0.1 M urea and 0.1 M NaCl (b) 0.1 M urea and 0.2 M MgCl2
(c) 0.1 M NaCl and 0.1 M Na2SO4 (d) 0.1 M Ca (NO3)2 and 0.1 M Na2SO4

24. When a mango fruit is kept in a very dilute solution of HCl, the fruit size
a. increase (b) decrease
(c) remains same (d) first decrease and then increases

25. Which of the following statements is not correct?

a. osmotic pressure is directly proportional to molar concentration
b. hypertonic solutions have lower concentrations with respect to reference
solution
c. isotonic solutions have same molar concentrations
d. osmotic pressure depends upon temperature

26. Osmotic pressure observed when benzoic acid is dissolved in benzene is less
than that expected from theoretical consideration. That is because
a. benzoic acid has higher molar mass than benzene
b. benzoic acid an organic solute
c. benzoic acid gets dissociated in benzene
d. benzoic acid gets associated in benzene

27. van’t Hoff factor (i) for aqueous solutions of electrolytes is

a. zero (b) greater than 1
(c) less than one (d) depends on nature of electrolyte

28. Mention one practical utility of reverse osmosis. (March 2022)

Ans. Desalination of sea water.

29. At constant temperature, different gases have different KH value. What does
this statement suggest? (March 2022)
Ans. KH depends on the nature of the gas OR Solubility of gas in liquid OR Different
gases will have different solubility.

30. What is the value of Van’t Hoff factor (i) for K2SO4? (March 2020)
Ans. 3

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31. 10 mL of liquid A is mixed with 10 mL of liquid B, the volume of the resultant
solution is 19.9 mL. What type of deviation is expected from Raoult’s s law?
(March 2020)
Ans. Negative deviation from Raoult’s s law

32. State Henry’s law. ( March 2018)

Ans. Henry’s law states that partial pressure of the gas in vapour phase is
proportional to mole fraction of the gas in the solution.
p = KH x χ
Where p is the partial pressure of the gas over the solution
χ is the mole fraction and K H is is Henry’s constant.

33. Van’t Factor for solution is less than one, what is the conclusion drawn from
it?
Ans. Solute undergoes association.

34. How molarity does varies with temperature? ( March 2017 )

Ans. Molarity decreases with increase in temperature, or Molarity increases with
decrease in temperature, or Molarity varies inversely with temperature.

35. 10 mL of liquid 'A' is mixed with 10 mL of liquid 'B', the volume of the
resultant solution is 19.9 mL. What type of deviation expected from Raoult's
law?
Ans. Non ideal solution with negative deviation

36. Name the law behind dissolution of CO2 in soft drinks under high pressure.
(March 2016)
Ans. Henrys law.

37. Ornamental gold containing copper is example for what type of solution?
(March 2016)
Ans. Solid solution.

38. Soda water bottles are sealed under high pressure. Give reason. (June 2016)
Ans. Increase the solubility of carbon dioxide.

39. What are ideal solutions? (June 2016)

Ans. Intermolecular force of attraction between two components and force of
attraction between the individual components are same.

40. What is the effect of temperature on the solubility of gases in liquids ? ( June
2016 )
Ans. Increase in temperature decreases the solubility of gases.

41. At a given temperature and pressure nitrogen gas is more soluble in water
than Helium gas. Which one of them has higher value KH? ( March 2015 )
Ans. Helium.

42. On mixing equal volumes of acetone and ethanol, what type of deviation from
Raoult’s law is expected?
Ans. Positive deviation.
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43. On what factor does the value of colligative property depend? ( July 2015 )
Ans. number of solute particles.

44. Give an example for liquid solution in which solute is gas. ( July 2015 )
Ans. Carbonated water.

45. At a given temperature and pressure nitrogen gas is more soluble in water
than Helium gas. Which one of them has higher value KH? ( March 2015 )
Ans. Helium.

46. On mixing equal volumes of acetone and ethanol, what type of deviation from
Raoult’s law expected? ( March 2015 )
Ans. Positive deviation.

47. What is Binary Solution? ( July 2014 )

Ans. Solution which contain two components is called binary solution.

48. Define Molality. ( July 2014 )

Ans. Molality is the defined as the number of mole of solute dissolved in one Kg of
the solvent.

49. Define the term ‘molarity’? ( March 2014 )

Ans. Molarity is defined as the number of moles of solute dissolved in one dm3 or
one liter of the solution.

50. Mention the value of enthalpy of mixing (∆𝑯𝒎𝒊𝒙) to form an ideal solution.
(March 2014)
Ans. Zero.

51. The experimental value for the molar mass of a non-volatile solute is twice the
theoretical value. What is the Van’t Hoff factor for the solute? (MQP)
Ans. Half or 0.5

52. Define ppm.(MQP)

Ans. Parts per million (ppm) is defined as the number of parts of mass the solute
per million parts by mass the solution. OR
Mass of the solute
PPM = Mass of the solution x 106

53. Give an example for a solution of a solid in a gas.

Ans. Camphor in oxygen.

54. 5 g of glucose is dissolved in 95g of water. What is the mass percentage of

glucose?
Ans.
mass of solute
Mass percentage = x 100
mass of solution

5
Mass percentage = x 100 = 5 %
95 + 5
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55. Write the expression to calculate volume percentage of solute.
Ans.
Volume of solute
Volume percentage = x 100
Volume of solution

56. In a binary solution, mole fraction of a component is 0.85. What is the mole
fraction of the other component?
Ans. 1 – 0.85 = .15.

57. What is the mass of sodium hydroxide present in 500mL of 1M NaOH

solution? (Molar mass of NaOH is 40 gmol-1).
Ans. 20 gm

58. Name a concentration term which is independent of temperature.

Ans. Molality.

59. How does the solubility of a solid solute in a liquid vary with increase in
temperature if the dissolution process is endothermic?
Ans. Solubility Increases.

60. Write the mathematical expression for Henry's law.

Ans. p = K H x χ

61. KH values for the gases argon and methane in water at 298K are 40.3 k bar and
0.413 k bar respectively. Which gas is more soluble at this temperature?
Ans. Methane.

62. Cylinders used by Scuba divers is diluted with helium gas. Why?
Ans. Because helium insoluble in blood.

63. Vapour pressures of chloroform and dichloromethane are 200mm of Hg and

415mm of Hg at 298K respectively. Which one is more volatile?
Ans. Dichloromethane.

64. Give an example for a non ideal solution showing negative deviation from
Raoult’s law.
Ans. Acetone and chloroform.

65. Based on inter molecular interactions, give the reason for a solution of A and
B to show positive deviation from Raoult’s law.
Ans. Intermolecular interactions between A and B are lesser than the
intermolecular interactions between the A – A and B – B.

66. Arrange the following aqueous solutions in decreasing order of their relative
lowering of vapour pressure:
i. 0.1M sucrose ii. 0.1 M NaCl iii. 0.05 M glucose iv. 0.1 M acetic acid.
Ans. 0.1 M NaCl ii. 0.1 M Sucrose iii. 0.05 M glucose iv. 0.1 M acetic acid

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67. Molal elevation constant for water is 0.52 K kg mol-1. What is the elevation in
boiling point produced for one molal aqueous solution of a solute for which
i=1?
Ans.

ΔTb = Kb × m × i

= 0.52 ×1×1= 0.52;

∴Tb =100 + 0.52=100.52°C

68. Which of the following aqueous solutions should have higher boiling point?
0.01M NaCl or 0.01M Na2SO4 (assume both undergo almost complete
ionisation)
Ans. 0.01M Na2SO4

69. Sea water freezes at a temperature lower than that of pure water. Why?
Ans. Because the sea water contains electrolytes.

70. Ethylene glycol is added as antifreeze to petrol in cold countries. Why?

Ans. Because it decreases the freezing point.

71. Which solution would exhibit lower osmotic pressure? Aqueous solution of
urea or aqueous solution of common salt, both of same concentration.
Ans. Aqueous solution of common salt.

72. Give a definition for Van’t Hoff factor 'i'.

Ans.
total number of particles after dissociation/association
i=
number of particles before dissociation/association

73. Write the SI unit for Ebullioscopic constant.

OR
What is the S.I unit of molal elevation constant ( kb) of a solvent ?
Ans. Kg K mol-1

74. 68% aqueous nitric acid cannot be concentrated by further fractional

distillation. Give reason.
Ans. Because they form Azeotropic mixtures. Or they boil at same temperature.

75. The cryoscopic constant and freezing point of benzene is 5.12 K kg mol-1 and
278.6 K respectively. At what temperature will one molal solution of benzene
containing a nonelectrolyte (i=1) freeze?
Ans. 273.48.

76. Name the phenomenon involved: A raw mango in a concentrated salt solution
loses water and shrinks.
Ans. Osmosis.

77. How does the solubility of a solute vary with increase in temperature if the
dissolution process is exothermic?
Ans. Decreases.
16
78. What are Solid solutions?
Ans. Mixture of solid as solvent and solid, gas, or liquid as solute.

79. What are colligative properties?

Ans. Colligative properties of a dilute solution containing a non-volatile solute are
those properties whose values depend only on the number of solute particles
present in the solution and not on its nature, size or chemical composition.

80. Name a colligative property.

Ans. Osmotic pressure, lowering of vapour pressure, elevation in boiling point, and
depression freezing point. ( any one of the above)

81. What does the Van’t Hoff factor ‘i’ for a solute in a solvent account for?
Ans. Extent of association or dissociation.

82. What is the effect of increase in pressure on the solubility of a gas in a liquid ?
Ans. Increase in pressure increases the solubilities.

83. Components of a non-ideal binary solution cannot be completely separated by

fractional distillation. Why?
Ans. Because they form azeoptropic mixtures.

84. What is the Van’t Hoff factor for potassium sulphate in very dilute aqueous
solution?
Ans. Three.

85. What happens to vapour pressure of water if a table spoon of sugar is added to
it ?
Ans. Vapour pressure decreases.

86. Two liquids A and B boils at 145oc and 190oc respectively, which of them has
higher vapour pressure at 80oc?
Ans. Liquid A

87. What will happen if pressure greater than osmotic pressure is applied on the
solution separated by semipermeable membrane from the solvent ?
Ans. Reverse osmosis takes place.

88. State Raoult’s law of a solution containing non-volatile solute.

Ans. The relative lowering of vapour pressure is equal to the mole fraction of the
solute present in the solution.
PA0 − P
= 𝜒𝐵
PA0

89. Van’t Hoff factor ‘i’ is greater than 1 for aqueous solution containing
potassium chloride. Why?
Ans. Because KCl undergoes dissociation.

17
90. “Aquatic animals are more comfortable in cold water rather than in warm
water” ?
Ans. Because solubility of oxygen is more in cold water than in warm water.

91. What is hypertonic solution?

Ans. A solution which has higher osmotic pressure than the other solution is called
hypertonic solution.

92. What is Azeotropic mixture?

Ans. Mixture of two liquids having same boiling point are called Azeotropic mixture.

93. A non-ideal solution has ∆H mixing > 0. What type of deviation does it show
from Raoult’s law?
Ans. Positive deviation.

Questions carrying two mark each

1. Define Van’t Hoff factor (i). give the value of ‘I’ for complete dimerization of all
the molecules of ethanoic acid in benzene. (June 2023)
Ans.
total number of particles after dissociation/association
i=
number of particles before dissociation/association

Van’t Hoff’s factor = i = 2

2. Define molarity (M). How does molarity vary with temperature? (May 2022)
Ans. The number of moles of solute dissolved in 1 liter (1 dm3) or 1000 cm3 of a
solution is called molarity.
Molarity varies inversely with temperature. OR As the temperature increases,
molarity of the solution decreases.

3. What type of deviation from Raoult’s law is observed When equal volume of
ethanol and acetone are mixed? Mention the reason for it. (May 2022)
Ans. Positive deviation from Raoult’s law.
Because
1. Force of attraction between ethanol and acetone molecules is less than that of
ethanol – ethanol & acetone – acetone molecules. OR
2. The ethanol-acetone shows weaker interaction than pure ethanol-ethanol
interaction.

4. What type of deviation from Raoult’s law is observed When equal volume of
ethanol and acetone are mixed? Mention the reason for it.
Ans. +ve deviation OR Positive deviation from Raoult’s law.
Because
1. Force of attraction between ethanol and acetone molecules is less than that of
ethanol – ethanol & acetone – acetone molecules. Or
2. Attractive Force between A - B molecules is less than that of A - A & B – B
molecules.
3. In pure ethanol molecules are hydrogen bonded. On adding acetone, its
molecules get in between the ethanol molecules and break hydrogen bonds between
18
 them. Or
4. The ethanol-acetone shows weaker interaction than pure ethanol-ethanol
interaction. (Any one answer)

5. What are azeotropes? Give an example for binary solutions showing minimum
boiling azeotrope. (March 2020)
Ans. Binary liquid mixtures having the same composition in liquid and vapour
phase and boil at a constant temperature.
Or
Constant boiling point liquid mixture (Any suitable example for solution showing
minimum boiling azeotrope.)
benzene and acetone, n-Hexane and ethanol, water and ethanol, acetone and CS2,
CCl4 and CHCl3, CCl4 and Toluene, Acetone and ethanol.

6. Define
i. Molality. ( July 2014 )
ii. isotonic solutions? (March 2018)
Ans. i. Molality is the defined as the number of mole of solute dissolved in one Kg
of the solvent
ii. Solutions having same osmotic pressure at given temperature.

7. How solubility of a gas in liquid varies with (i) Temperature and (ii) pressure? (
March 2017)
Ans. Temperature: Increase in temperature decreases the solubility
Pressure: Increase in pressure increases the solubility.

8. What are non-ideal solutions? Mention the reason for the negative deviation
from Raoult’s law. ( March 2016 )
Ans. An non –ideal solution is one where the intermolecular interactions between
the components are different from the intermolecular interactions between the
molecules of the pure components.

9. Define isotonic solutions. What happens when blood cells is dipped in solution
containing more than normal saline concentration. (March 2015)
Ans. The solutions having same osmotic pressure at a given temperature is called
isotonic solutions. When the blood cells is dipped in solution containing more than
normal saline concentration, blood cells shrink.

10. What happens to the solubility of a gas in a liquid with increase in

temperature? Give reason. (July 2014)
Ans. Solubility decreases, because dissolution of gas in liquid is exothermic process.

11. What is reverse osmosis? Mention any one of it’s use.

Ans. When a pressure which is greater than osmotic pressure is applied on the
solution, the solvent molecules flow from the solution into the solvent through semi
permeable membrane. This process is called reverse osmosis.
Use: used in desalination of water.

12. A 4% solution of a non –volatile solute is isotonic with 0.702 % urea solution.
Calculate the molar mass of the non – volatile solute. (Molar mass of urea=60 g
mol-1)
19
 Ans.
W𝐀 W𝐁
=
MA MB

40 x 60
MB = = 341.8 g
0.702

13. State Raoult’s law for a solution of 2 volatile liquids. Give an example for
liquid mixture that shows negative deviation from Raoult’s law?
Ans. According to Raoult’s law, At a given temperature the partial pressure of each
component in solution is equal to the product of the vapour pressure of the pure
component and its mole fraction. PA = PA0 x χA and PB = PB0 x χB

Solutions showing negative deviation: Acetone + chloroform or Nitric acid + water

14. Point out the difference between ideal and non ideal solution. ( June 2015,
March 2016 )
Ans.

Ideal solution Non-ideal solution

Obeys Raoult’s law Does not obeys Raoult’s law
∆𝑉𝑚𝑖𝑥 = 0 ∆𝑉𝑚𝑖𝑥 ≠ 0
∆𝐻𝑚𝑖𝑥 = 0 ∆𝐻𝑚𝑖𝑥 = 0

15. State Henry’s law. Write its mathematical form. ( March2014 ) ( June 2016 )
Ans. Henry’s law states that partial pressure of the gas in vapour phase is
proportional to mole fraction of the gas in the solution.
p = KH x χ
Where p is the partial pressure of the gas over the solution
χ is the mole fraction and K H is is Henry’s constant.

16. Calculate the mole fraction of CO2 in one litre of soda water sealed under a
pressure of 3.5 bar at 298 K. KH = 1.67 x 103 bar.
Ans. p = K H x χ

p
χ=
KH

3.5
χ= = 2.09 𝑥 10−3
1.67 x 103

17. Write any Two applications of Henry’s law.

Ans. To increase the solubility of carbon dioxide in soda water, soft drinks, the
containers sealed under high pressure.
At high altitudes, the partial pressure of oxygen is low. As a result, concentration of
oxygen becomes low in blood. Hence people become weak and the thinking ability
reduces. These symptoms are called anoxia.

18. What are azeotropes? Give an example.

Ans. Mixture of two liquids having same boiling point are called Azeotropic mixture.
Example: Water + ethyl alcohol
20
19. The vapour pressure of ethyl alcohol at 298k is 40mm of Hg. Its mole fraction
in a solution with methyl alcohol is 0.80. What is its vapour pressure in
solution if the mixture obeys Raoult’s law?
Ans. PA = PA0 x χA

PA = 40 x 0.80 = 32 mm of Hg

Questions carrying three mark each

20. 1.0 g of non-electrolyte solute dissolved in 50 g of benzene lowered the

freezing point of benzene by 0.4 K. Find the molar mass of the solute. [Given:
Freezing point, depression constant of benzene = 5.12 K. kg mol-1]. (June
2023, March 2017)
Ans.

1000 K f WB
ΔTf =
WA MB

1000 x 5.12 x 1.0

MB = = 256 g mol−1
50 x 0.4

21. 450 cm3 of an aqueous solution of a protein contains 1.0g of the protein. The
osmotic pressure of such a solution at 310K is found to be 3.1 X 10−4 bar.
Calculate the molar mass of the protein. (R = 0.083Lbar mol-1 K-1). (March
2023)
Ans.
WB R T
Π =
MB V

1.0 x 0.0823 x 310

MB = = 1,84,444 g mol−1
3.1 x 10−4 x 0.450

22. 31 g of an unknown material is dissolved in 500g of water. The resulting

solution freezes at 271.14 K. Calculate the molar mass of the material.
[Given: Kf for water = 1.86K Kgmol-1, Tf0 of water = 273K](July 2022)
Ans.
1000 K f WB
ΔTf =
WA MB
ΔTf = 𝑇𝑓0 − 𝑇𝑓 = 273 − 271.14 = 1.86
1000 x 1.86 x 31
MB = = 62 g mol−1
500 x 1.86

23. On dissolving 3.46g of solute in 100g of water, the boiling point of solution
was raised to that of pure water by 0.12K. Calculate the molar mass of the
non–volatile solute. (Given: Kb of water is 0.51 K kg mol-1)
1000 Kb WB
Ans. Δ𝑇𝑏 = W A MB

21
 1000 𝑥 0.51𝑥 3.46
MB =
0.12 x 100
g
MB = 147.5
mol

24. Vapour pressure of benzene is 200 mm of Hg. When 2 gram of a non-volatile

solute dissolved in 78 gram benzene. Benzene has vapour pressure of 195 mm
of Hg. Calculate the molar mass of the solute. [molar mass of benzene is 78
gram mol-1] (March 2020)
Ans.
p0A − p WB MA
0 =
pA WA MB

200 − 195 2 x 78
=
200 MB x 78

2 x 78 x 200
MB = = 80 𝑔 𝑚𝑜𝑙 −1
78 x 5

25. The boiling point of benzene is 353.23K. When 1.80 g of a non- volatile solute
was dissolved in 90 g of benzene, the boiling point is raised to 354.11K.
Calculate the molar mass of the solute (kb= 2.53K Kg mol-1) (March 2018, June
2015)
Ans.
K b 1000 WB
Δ𝑇𝑏 =
WA MB

K b 1000 WB
MB =
WA Δ𝑇𝑏

2.53 x 1000 x 1.8

MB =
0.88 x 90
g
MB = 57.5
mol

26. 5.8 g of non-volatile solute was dissolved in 100 g of carbon disulphide. The
vapour pressure of the solution was found to be 190 mm of Hg. Calculate the
molar mass of the solute given the vapour pressure of pure carbon disulphide
is 195 mm of Hg. Molar mass of carbon disulphide is 76 g mol-1 ( March 2016 )
Ans.

22
 p0A − p WB MA
0 =
pA WA MB

195 − 190 5.8 x 76

=
195 100 x MB

5.8 x 76 x 195
MB = = 171.9 𝑔
100 x 5

27. 24 g of a non–volatile, non–electrolyte solute is added to 600 g of water. The

boiling point of the resulting solution is 373.35K. Calculate the molar mass of
the solute (Given boiling point of pure water = 373 K and Kb for water = 0.52 K
kg mol-1).
Ans.
1000 K b WB
Δ𝑇𝑏 =
WA MB

1000 K b WB
MB =
WA Δ𝑇𝑏

. 52 x 1000 x 24
MB =
600 x .35

MB = 59.42 g

28. 300 Cm3 of an aqueous solution of protein contains 2.12 g of the protein, the
osmotic pressure of such a solution at 300 K is found to be 3.89 x 10-3 bar.
Calculate the molar mass of the protein. ( R = 0.0823 L bar mol-1K-1 ) ( July
2016 )
Ans.

WB R T
Π =
MB V

2.12 x 0.0823 x 300

MB = = 44,852 g mol−1
3.89 x 10−3 x 0.3

29. 200 Cm3 of an aqueous solution of protein contains 1.26 g of the protein, the
osmotic pressure of such a solution at 300 K is found to be 2.57 x 10-3 bar.
Calculate the molar mass of the protein. ( R = 0.0823 L bar mol-1K-1 ) ( March
2016 )
Ans.
WB R T
Π =
MB V
1.26 x 0.0823 x 300
MB =
2.57 x 10−3 x 0.2

MB = 60,039 g mol−1

23
30. A solution containing 18 g non-volatile non-electrolyte solute is dissolved in
200 g of water freezes at 272.07 K. Calculate the molecular mass of solute.
Given Kf = 1.86 k Kg / mol. Freezing point of water = 273 K ( March 2015 )
Ans.
1000 K f WB
ΔTf =
WA MB

1000 x 1.86 x 18
MB =
200 x (273 − 272.07)

1000 x 1.86 x 18
MB =
200 x (273 − 272.07)

MB = 180 g mol−1

31. On Dissolving 2.34 g of solute in 40 g benzene, the boiling point of solution

was higher than that of benzene by 0.81 K. Kb value of benzene is 2.5 K kg
mol-1 . Calculate the molar mass of the solute. ( March 2014 )
Ans.
K b 1000 WB
Δ𝑇𝑏 =
WA MB

K b 1000 WB
MB =
WA Δ𝑇𝑏

2.5 x 1000 x 2.34

MB =
0.81 x 40

MB = 180.5 g

32. The vapour pressure of pure benzene at a certain temperature is 0.850 bar.
When 0.5 g of a non-volatile solute is added to 39.0 g of benzene ( Molar mas
of benzene =78 g mol-1) Vapour pressure of the solution is 0.845 bar. What is
the molar mass of a non-volatile solute? ( July 2014 )
Ans.
p0A − p WB MA
0 =
pA WA MB

0.85 − 0.845 0.5 x 78

=
0.85 MB x 39

MB = 170 g mol−1

33. Plot a graph of vapour pressure against mole fraction of the two volatile
liquids forming an ideal solution. What is the change in enthalpy upon mixing
the two components of an ideal solution?
Ans.

24
 o
pB

o
pA

pressure
Vapour
pressur
Vapour

χA = 1 χA = 0
Mole Fraction
χB = 0 χB = 1

34. 15.0 g of unknown substance was dissolved in 450 g of water. The resulting
solution was found to freeze at -0.34o C. Calculate the molar mas of the
substance.
Ans.
K f 1000 WB
MB =
WA Δ𝑇𝑓

1.86 x 1000 x 15
MB = = 182.35 g mol−1
450 x 0.34

35. If 1.71 g of sugar (Molar mass= 342) is dissolved in 500 cm3 of a solution at
300k. What will be its osmotic pressure? (Given R = 0.0831 L bar k-1 mol-1)
Ans.
WRT
Π =
MB V

1.71 x 0.0831 x 300

Π = = 0.249 bar
343 x 0.5

36. At 298k, the vapour pressure of a water is 23.75mm of Hg. Calculate the
vapour pressure at the same temperature over 5% aqueous solution of urea.
given the molecular mass of urea= 60
Ans.
p0A − p WB MA
=
p0A WA MB

23.75 − P 5 x 18
=
23.75 95 x 60

5 x 18 x 23.75
23.75 − P =
95 x 60

23.75 − P = 0.375

P = 23.75 − 0.375 = 23.37 mm of Hg

37. For a non-ideal solution having positive deviation from Raoult’s law.
i. Plot a graph of vapour pressure against mole fraction.
ii. What type of azeotropes formed by this solution?

25
 iii. Give one example for the above solution.
Ans.

o
pB
pressure
Vapour

o
pA

pressure
Vapour
χA = 1 Mole χA = 0
χB = 0 Fraction χB = 1
ii. Minimum boiling azeotrope is formed.
iii. Ethanol + Acetone

38. Vapour pressure of liquids A and B at 298k is 300mm of Hg and 450mm of Hg,
calculate the mole fraction of A in the mixture.
Ans. Let the mole fraction of A be XA
Let the mole fraction of B be XB
XB = 1 – XA
From Raoult’s law,
Ptotal = PoXA + PoXB
Ptotal = PoXA + Po ( 1 - XB )
405 = 300 XA + 450 ( 1 - XA )
XA = 0.3

26
Questions from previous year PU Board examinations

UNIT-1 Solution
March -2023
1. A binary liquid mixture that forms maximum boiling azeotrope as a specific
composition is
(a) Ethanol + Water (b) n-Hexane + n-Heptane
(c) Benzene + Toluene (d) Nitric Acid + Water
2. The Value of Van’t Hoff Factor (i) for ethanoic acid in benzene is nearly
(a) 2 (b) 1
(c) 0.5 (d) 0
3. Because of low concentration of O2 in the blood and tissues of people living at high
altitudes, suffe from a disease called_______________
4. a) 450 cm3 of an aqueous solution of a protein contains 1.0 g of the protein. The
osmotic pressure such a solution at 310 K is found to be 3.1 x 10-4 bar. Calculate
the molar mass of the protein(R = 0.083 L bar mol-1 K-1)
b) State Raoult’s law of relative lowering of vapour pressure. Write its mathematical
form.

March 2022
1. Mention one practical utility of reverse osmosis.
2. At Constant temperature, different gases have different KH – value. What does the
statement suggest?
3. Define Molarity [M]. How does molarity vary with temperature?
4. a) On dissolving 3.46 g of non-volatile solute in 100 g of water, the boiling point of
solution was raised to that of pure water by 0.12 K. Calculate the molar mass of
non-volatile solute. (Given: Kb of water = 0.51 K Kg mol-1).
b) What type to deviation from Raoult’s law is ordered when volume of ethanol equal
and acetone are mixed? Mention the reason for it.

July-2022
1. State Henry’s law.
2. Sodium chloride solution having concentration greater than 0.9% is not sage to
inject intravenously- Give reason.
3. Give any two difference between ideal and non-ideal solutions.
4. a. 31 g of an unknown material is dissolved in 500g of water. The resulting solution
freezes at 271.14 K. Calculate the molar mass of the material.
b. (i) What are azeotropes?
(ii) Which type of azeotrope shows a large positive deviation from Raoult’s law?

March 2020
1. What is the value of Van’t Hoff factor (i) for K2SO4?
2. 10 mL of liquid A is mixed with 10 mL of liquid B, the volume of the resultant
solution is 19.9 mL, what type of deviation is expected from Raoult’s law?
3. a. Vapour pressure of benzene is 200 mm of Hg. When 2 gram of a non-volatile
solute dissolved in 78 gram benzene, benzene has vapour pressure of 195 mm of
Hg. Calculate the molar mass of the solute. [molar mass of benzene is 78 gram mol-
1]

27
 b. What are azeotropes? Give a example for binary solutions showing minimum
boiling azeotrope.

July 2020
1. Define Molality.
2. The KH values for Nitrogen gas (N2) at 293 K and 303 K are 76.48 Khar and 88.84
Khar respectively. Among these two given temperatures, at which temperature
Nitrogen gas is more soluble in water?
3. a. 1.00 g of a non-electrolyte solute is dissolved in 50g of benzene lowers the
freezing point of benzene by 0.4K. The freezing point depression constant for
benzene is 5.12 K Kg mol-1. Find the molar mass of the solute.
b. State Henry’s law. Write its mathematical form.

March 2019
1. How does the size of blood cells change when placed in an aqueous solution
containing more than 0.9% (m/v) sodium chloride?
2. How does the volume change on missing two volatile liquids to form an ideal
solution?
3. a. 31g of an unknown molecular material is dissolved in 500g of water. The
resulting solutions freezes at 271.14 K. Calculate the molar mass of the material.
[Given : Kf for water = 1.86 Kkgmol-1, Tf0 of water = 273 K].
b. What is reverse osmosis? Mention its use.

July 2019
1. Write the unit of molality of a solution.
2. At a given temperature oxygen gas is more soluble in water than Nitrogen gas.
Which one of them has higher value of KH?
3. a. On dissolving 2.34 g of non-electrolyte solute in 40g of benzene, the boiling point
of solution was higher than benzene by 0.81 K. Kb value for benzene is 2.53 K kg
mol-1. Calculate the molar mass of solute. [Molar mass of benzene is 78g mol-1].
b. (i) State Henry’s law.
(ii) How solubility of a gas in liquid changes with increase in temperature?

March 2018
1. State Henry’s law.
2. Van’t Hoff’s factor for a solution is less than one, what is the conclusion drawn from
it?
3. a. The boiling point of benzene is 353.23 K. When 1.80g of a non-volatile, non-
ionisable solute was dissolved in 90g of benzene, the boiling point raised to 354.11
K. Calculate molar mass of the solute. [Kb for benzene = 2.53 K Kg mol-1].
b. Define (i) Molality of a solution. (ii) Isotonic solutions.

March 2018
1. In a binary solution, mole fraction of one component is 0.068. what is the mole
fraction of another component?
2. State Henry’s law.
3. a. 5.8g of non-volatile, non-electrolyte solute was dissolved in 100g of carbon
disulphide (CS2). The vapour pressure of the solution was found to be 190 mm of
Hg. Calculate molar mass of the solute. Given : Vapour pressure of pure CS2 is 195
mm of Hg and molar mass of CS2 is 76g/mol.
b. Mention any two differences between ideal and non-ideal solutions.
28
 March 2017
1. How molarity does varies with temperature?
2. 10 mL of liquid ‘A’ is mixed with 10 mL of liquid ‘B’ the volume of the resultant
solution is 19.9 mL. what type of deviation expected from Raoult’s law?

3. A. 1.0 g of non-electrolyte solute dissolved in 50 g of benzene lowered the freezing

point of benzene by 0.4 K. Find the molar mass of the solute.[Given: Freezing point
depression constant of benzene = 5012 K. kg mol-1].
b. How solubility of a gas in liquid varies with (i) Temperature and (ii) pressure?

July 2017
1. State Raoult’s law of a binary solution for two volatile liquid components.
2. Van’t Hoff factor for a solution is more than one. What is the conclusion drawn
from it?
3. a. The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A
non-volatile, non-electrolyte solid weighting 0.5 g, when added to 39.0 g of benzene
(molar mass of benzene 78 g mol-1) vapour pressure of the solution then is 0.845
bar. What is the molar mass of the solid substances?
b. What is Reverse Osmosis? Mention its one practical utility.

March 2016
1. Name the law behind the dissolution of CO2 gas in soft drinks under high pressure.
2. Ornamental gold, containing copper is an example for what type of solution?
3. a. 5.8g of a non volatile solute was dissolved in 100 g of carbon disulphide (CS2).
The vapour pressure of the solution was found to be 190 mm. of Hg. Calculate the
molar mass of the solute given the vapour pressure of pure CSs in 195 mm. of Hg
[Molar mass of CS2 = 76 g mol-1]
b. Mention any two differences between Ideal and non ideal solutions.

July 2016
1. What are isotonic solutions.
2. What is the effect of rise in temperature on the solubility gases in liquids?
3. a. 300 cm3 of an aqueous solution of a protein contains 2.12 g of the protein, the
osmotic pressure of such a solution at 300 K is found to be 3.89 x 10-3 bar.
Calculate the molar mass of the protein. (R = 0.0823 L bar mol-1 K-1)
b. (i) State Henry’s Law
(ii) Soda water bottles are sealed under high pressure. Give reason.

March 2015
1. At a given temperature and Pressure nitrogen gas is more soluble in water than
Helium gas. Which one of them has higher value of Kn?
2. On mixing equal volumes of acetone and ethanol, what type of deviation from
Raoult’s law expected?
3. a. A solution containing 18g of non-volatile non-electrolyte solute is dissolved in 200
g of water freeze at 272.07K. Calculate the molecular mass of solute. Given K1 =
1.86 K kg/mol. Freezing point of water = 273 K.
b. Define isotonic solution. What happens when the blood cell is dipped in a solution
containing more than normal saline concentration?

29
 July 2015
1. On what factor the value of colligative property depends?

2. Give an example for liquid solution in which solute is gas.

3. a. The boiling point of benzene is 353.23 K When 1.80 g of a non-volatile, non-
ionising solute dissolved in 90 g of benzene, the boiling point is rated 354.11 K.
Calculate the molar mass of solute. [Given Kb for benzene = 2.53 K kg mol-1]

March 2014
1. Define the term “molarity’.
2. Mention the enthalpy of mixing (∆mix H) value to form an ideal solution.
3. a. On dissolving 2.34 g of solute in 40 g of benzene, the boiling point of solution was
higher than the benzene by 0.81 K. Kb value of benzene is 2.53 K kg mol-1. Calculate
the molar mass of the solute.
b. State Henry’s law. Write its mathematical form.

July 2014

1. What is Binary Solution?

2. Define Molarity.
3. a. The vapour pressure of pure benzene at certain temperature is 0.850 bar. A non-
volatile, non-electrolyte solid weighing 0.5 grams when added to 39 grams of
benzene (molar mass 78 grams), vapour pressure of the solution, then is 0.845
bars. What is the molar mass of the solid substance?
b. What happens to the solubility of a gas in a liquid with increase in temperature?
Give reason.

30