VELAMMAL BODHI CAMPUS, THANJAVUR

GRADE XII SOLUTION SUB :

CHEMISTRY

1MARK

1. Define the following term : Molality (m) (1/2, Delhi 2017, 1/5 AI 2014C)

2. Define the following term : Molarity (M) (1/2, Delhi 2017, 1/5 AI 2014)

3. Define the following term : Mole fraction (1/5 AI 2014 C, 1/2, Delhi 2012, AI
2012)

4. Calculate the molarity of 9.8% (w/W) solution of H2SO4 if the density of

the solution is 1.02 g mL–1. (Molar mass of H2SO4 = 98 g mol–1) (2/5,
Foreign 2014)

5. Differentiate between molarity and molality of a solution. How can we

change molality value of a solution into molarity value? (Delhi 2014C)

6. A solution of glucose (C6H12O6) in water is labelled as 10% by weight.

What would be the molality of the solution? (Molar mass of glucose = 180 g
mol–1) (2/5, AI 2013)

7. Differentiate between molarity and molality in a solution. What is the

effect of temperature change on molarity and molality in a solution? (2/5,
Delhi 2011, 2/5, AI 2011)

8.Differentiate between molarity and molality of a solution. Explain how

molarity value of a solution can be converted into its molality? (Foreign
2011)

9. A solution of glucose (molar mass = 180 g mol–1) in water is labelled as

10% (by mass). What would be the molality and molarity of the solution?
(Density of solution = 1.2 g mL–1) (3/5, AI 2014)

10. Why aquatic animals are more comfortable in cold water than in warm
water? (1/3, 2018)

11. Gas (A) is more soluble in water than gas (B) at the same temperature.
Which one of the two gases will have the higher value of KH (Henry’s
constant) and why? (1/2, AI 2016)
12. Explain the following : Henry’s law about dissolution of a gas in a liquid.
(1/5, AI 2012)

13. State the following : Henry’s law about partial pressure of a gas in a
mixture. (1/5, Delhi, AI 2011)

14. State Henry’s law. Why is air diluted with helium in the tanks used by
scuba divers? (2/5, 2020)

15. State Henry’s law. Calculate the solubility of CO2 in water at 298 K under
760 mm Hg. (KH for CO2 in water at 298 K is 1.25 × 106 mm Hg) (2020)

16. Give reasons for the following. (a) Aquatic species are more comfortable
in cold water than in warm water. (b) At higher altitudes people suffer from
anoxia resulting in inability to think. (AI 2019)

17. State Henry’s law and mention two of its important applications. (2/5, AI
2013C, 2012C)

18. Explain why aquatic species are more comfortable in cold water rather
than in warm water. (Delhi 2012C)

19. The partial pressure of ethane over a saturated solution containing 6.56
× 10–2 g of ethane is 1 bar. If the solution contains 5.0 × 10–2 g of ethane,
then what will be the partial pressure of the gas? (Delhi 2013C, AI 2012C)

20. If N2 gas is bubbled through water at 293 K, how many millimoles of N2

gas would dissolve in 1 litre of water? Assume that N2 exerts a partial
pressure of 0.987 bar. Given that Henry’s law constant for N2 at 293 K is
76.48 kbar. (AI 2012C)

21. Identify which liquid will have a higher vapour pressure at 90°C if the
boiling points of two liquids A and B are 140°C and 180°C, respectively. (One
word, 2020)

22. Define Raoult’s law. (1/5 AI 2014C)

23. State the following : Raoult’s law in its general form in reference to
solutions. (1/5, Delhi, 1/2, AI 2011)

24. State Raoult’s law for the solution containing volatile components. What
is the similarity between Raoult’s law and Henry’s law? (Delhi 2014, AI 2013)
25. State Raoult’s law for a solution containing volatile components. Name
the solution which follows Raoult’s law at all concentrations and
temperatures. (2/5, Foreign 2014)

26. State Raoult’s law. How is it formulated for solutions of non-volatile

solutes? (Delhi 2012C)

27. The vapour pressure of pure liquids A and B are 450 and 700 mm Hg
respectively, at 350 K. Find out the composition of the liquid mixture if total
vapour pressure is 600 mm Hg. Also find the composition of the vapour
phase. (3/5, AI 2013C)

28. What happens when acetone is added to pure ethanol? (1/2, 2020)

29. Define the following term : Ideal solution (1/2, Delhi 2017, 1/5, AI 2017C,
1/5, AI 2013, 2012, 1/2, Delhi 2012)

30. In non-ideal solution, what type of deviation shows the formation of

maximum boiling azeotropes? (1/2, AI 2016)

31. Some liquids on mixing form ‘azeotropes’. What are ‘azeotropes’? (Delhi
2014)

32. Define the following term : Azeotrope (1/5, Foreign 2014) SA (2 marks)

33. Write two differences between ideal solutions and non-ideal solutions.
(Delhi 2019, 2/5, AI 2017)

34. What type of azeotropic mixture will be formed by a solution of acetone

and chloroform? Justify on the basis of strength of intermolecular interactions
that develop in the solution. (AI 2019)

35. What is meant by positive deviations from Raoult’s law? Give an

example. What is the sign of DmixH for positive deviation? (Delhi 2015)

36. Define azeotropes. What type of azeotrope is formed by positive

deviation from Raoult’s law? Give an example. (Delhi 2015)

37. What is meant by negative deviation from Raoult’s law? Give an

example. What is the sign of DmixH for negative deviation? (Foreign 2015)

38. Define azeotropes. What type of azeotrope is formed by negative

deviation from Raoult’s law? Give an example. (Foreign 2015)
39. What type of deviation is shown by a mixture of ethanol and acetone?
Give reason. (2/5, AI 2014)

40. What is meant by positive and negative deviations from Raoult’s law and
how is the sign of DmixH related to positive and negative deviations from
Raoult’s law? (AI 2013 C)

41. Explain why a solution of chloroform and acetone shows negative

deviation from Raoult’s law. (2/5, Delhi 2011C)

42. Assertion (A) : Osmotic pressure is a colligative property. Reason (R) :

Osmotic pressure is directly proportional to molarity.

(a) Both Assertion (A) and Reason (R) are correct statements, and Reason (R)
is the correct explanation of the Assertion (A).

(b) Both Assertion (A) and Reason (R) are correct statements, but Reason (R)
is not the correct explanation of the Assertion (A).

(c) Assertion (A) is correct, but Reason (R) is incorrect statement.

(d) Assertion (A) is incorrect, but Reason (R) is correct statement. (2020)

43. What happens when a pressure greater than osmotic pressure is applied
on the solution side separated from solvent by a semi-permeable
membrane? (1/2, 2020)

44. Give reason for the following : Measurement of osmotic pressure method
is preferred for the determination of molar masses of macromolecules such
as proteins and polymers. (1/2, 2018)

45. Define the following term : Colligative properties (1/2, Delhi 2017)

46. Define the following term : Osmotic pressure (1/5, AI 2017C, 2013)

47. What are isotonic solutions? (Delhi 2014; 1/2, Delhi 2012)

48. Define the following term : Molal elevation constant (Kb) (1/5, AI 2014)

49. How is the vapour pressure of a solvent affected when a non-volatile

solute is dissolved in it? (1/2, Delhi 2014C)

50. Explain the following : Boiling point elevation constant for a solvent. (AI
2012)
51. For a 5% solution of urea (Molar mass = 60 g/mol), calculate the osmotic
pressure at 300 K.[R = 0.0821 L atm K–1 mol–1] (2020)

52. Visha took two aqueous solutions – one containing 7.5 g of urea (Molar
mass = 60 g/mol) and the other containing 42.75 g of substance Z in 100 g
of water, respectively. It was observed that both the solutions froze at the
same temperature. Calculate the molar mass of Z. (2020)

53. Give reasons : (i) 0.1 M KCl has higher boiling point than 0.1 M glucose.
(ii) Meat is preserved for a longer time by salting. (2020)

54. Calculate the freezing point of a solution containing 60 g of glucose

(molar mass = 180 g mol–1) in 250 g of water. (Kf of water = 1.86 K kg mol–
1) (2018) 55. (i) Out of 1 M glucose and 2 M glucose, which one has a higher
boiling point and why? (ii) What happens when the external pressure applied
becomes more than the osmotic pressure of solution? (2/5, Delhi 2016)

56. Blood cells are isotonic with 0.9% sodium chloride solution. What
happens if we place blood cells in a solution containing (i) 1.2% sodium
chloride solution? (ii) 0.4% sodium chloride solution? (2/5, Delhi 2016)

57. Why does a solution containing non-volatile solute have higher boiling
point than the pure solvent? Why is elevation of boiling point a colligative
property? (AI 2015)

58. Calculate the mass of compound (molar mass = 256 g mol–1) to be

dissolved in 75 g of benzene to lower its freezing point by 0.48 K. (Kf = 5.12
K kg mol–1). (Delhi 2014)

59. 18 g of glucose, C6H12O6 (Molar mass = 180 g mol–1) is dissolved in 1

kg of water in a sauce pan. At what temperature will this solution boil? (Kb
for water = 0.52 K kg mol–1, boiling point of pure water = 373. 15 K) (Delhi
2013)

60. An aqueous solution of sodium chloride freezes below 273 K. Explain the
lowering in freezing point of water with the help of a suitable diagram. (Delhi
2013C)

61. Define the terms osmosis and osmotic pressure. Is the osmotic pressure
of a solution a colligative property? Explain. (2/5, Delhi 2011)

62. List any four factors on which the colligative properties of a solution
depend. (2/5, AI 2011C)
63. Calculate the mass of ascorbic acid (Molar mass = 176 g mol–1) to be
dissolved in 75 g of acetic acid, to lower its freezing point by 1.5°C . (Kf = 3.9
K kg mol–1) (2020)

64. A 4% solution (w/w) of sucrose (M = 342 g mol–1) in water has a

freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M =
180 g mol–1) in water. (Given : Freezing point of pure water = 273.15 K)
(Delhi 2019)

65. At 300 K, 30 g of glucose present in a litre of its solution has an osmotic

pressure of 4.98 bar. If the osmotic pressure of a glucose solution is 1.52 bar
at the same temperature, what would be its concentration? (AI 2019)

66. A 10% solution (by mass) of sucrose in water has freezing point of 269.15
K. Calculate the freezing point of 10% glucose in water, if freezing point of
pure water is 273.15 K. (Given : Molar mass of sucrose = 342 g mol–1, molar
mass of glucose = 180 g mol–1) (Delhi 2017)

67. 30 g of urea (M = 60 g mol–1] is dissolved in 846 g of water. Calculate

the vapour pressure of water for this solution if vapour pressure of pure
water at 298 K is 23.8 mm Hg. (3/5, AI 2017)

68. Calculate the boiling point elevation for a solution prepared by adding 10
g of CaCl2 to 200 g of water. (Kb for water = 0.52 K kg mol–1, molar mass of
CaCl2 = 111 g mol–1) (3/5, AI 2017C, Foreign 2014)

69. Calculate the freezing point of the solution when 31 g of ethylene glycol
(C2H6O2) is dissolved in 500 g of water. (Kf for water = 1.86 K kg mol–1) (AI
2015)

70. A solution containing 15 g urea (molar mass = 60 g mol–1) per litre of

solution in water has the same osmotic pressure (isotonic) as a solution of
glucose (molar mass = 180 g mol–1) in water. Calculate the mass of glucose
present in one litre of its solution. (3/5, AI 2014)

71. Define the following terms : (i) Osmotic pressure (ii) Colligative
properties (Foreign 2014)

72. Some ethylene glycol, HOCH2CH2OH, is added to your car’s cooling

system along with 5 kg of water. If the freezing point of water-glycol
solution is –15.0°C, what is the boiling point of the solution? (Kb = 0.52 K kg
mol–1 and Kf = 1.86 K kg mol–1 for water) (Delhi 2014C)
73. 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered
the freezing point of benzene by 0.40 K. The freezing point depression
constant of benzene is 5.12 K kg mol–1. Find the molar mass of the solute.
(AI 2013)

74. A 5% solution (by mass) of cane-sugar in water has freezing point of 271
K. Calculate the freezing point of 5% solution (by mass) of glucose in water if
the freezing point of pure water is 273.15 K. [Molecular masses : Glucose
C6H12O6 : 180 amu; Cane-sugar C12H22O11 : 342 amu] (3/5, AI 2013C)

75. A solution of glycerol (C3H8O3) in water was prepared by dissolving

some glycerol in 500 g of water. This solution has a boiling point of 100.42°C
while pure water boils at 100°C. What mass of glycerol was dissolved to
make the solution? (Kb for water = 0.512 K kg mol–1) (Delhi 2012, AI 2012)

76. A solution containing 30 g of non-volatile solute exactly in 90 g of water

has a vapour pressure of 2.8 kPa at 298 K. Further 18 g of water is added to
this solution. The new vapour pressure becomes 2.9 kPa at 298 K. Calculate
(i) the molecular mass of solute and (ii) vapour pressure of water at 298 K.
(Delhi 2012C)

77. Calculate the boiling point of a solution prepared by adding 15.00 g of

NaCl to 250.00 g of water. (Kb for water = 0.512 K kg mol–1, Molar mass of
NaCl = 58.44 g) (3/5, Delhi 2011)

78. A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL

of water has an osmotic pressure of 0.335 torr at 25°C. Assuming the gene
fragment is a non-electrolyte, determine its molar mass. (3/5, Delhi, AI
2011)

79. What would be the molar mass of a compound if 6.21 g of it dissolved in

24.0 g of chloroform to form a solution that has a boiling point of 68.04°C.
The boiling point of pure chloroform is 61.7°C and the boiling point elevation
constant, Kb for chloroform is 3.63°C/m. (3/5, Delhi 2011)

80. The molecular masses of polymers are determined by osmotic pressure

method and not by measuring other colligative properties. Give two reasons.
(3/5, AI 2011C)

81. Calculate the boiling point of one molar aqueous solution (density 1.06 g
mL–1) of KBr. [Given : Kb for H2O = 0.52 K kg mol–1, atomic mass : K = 39,
Br = 80] (3/5, AI 2011C)
82.Give reason for the following : Elevation of boiling point of 1 M KCl
solution is nearly double than that of 1 M sugar solution. (1/2, 2018)

83. Define the following term : van’t Hoff factor (Delhi 2017, 1/5, Delhi 2012)
84. Define the following term : Abnormal molar mass (Delhi 2017)

85. What is van’t Hoff factor? What types of values can it have if in forming
the solution the solute molecules undergo (i) dissociation (ii) association?
(2/5, AI 2014C)

86. Assuming complete dissociation, calculate the expected freezing point of

a solution prepared by dissolving 6.00 g of Glauber’s salt, Na2SO4⋅10H2O in
0.100 kg of water. (Kf for water = 1.86 K kg mol–1, atomic masses : Na =
23, S = 32, O = 16, H = 1) (2/5, AI 2014C)

87. A 1.00 molal aqueous solution of trichloroacetic acid (CCl3COOH) is

heated to its boiling point. The solution has the boiling point of 100.18°C.
Determine the van’t Hoff factor for trichloroacetic acid. (Kb for water = 0.512
K kg mol–1) (Delhi 2012)

88. What is van’t Hoff factor? What possible value can it have if the solute
molecules undergo dissociation? (2/5, Delhi 2011C)

89. A solution contains 5.85 g NaCl (Molar mass = 58.5 g mol–1) per litre of
solution. It has an osmotic pressure of 4.75 atm at 27°C. Calculate the
degree of dissociation of NaCl in this solution. (Given : R = 0.082 L atm K–1
mol–1) (3/5, 2020)

90. When 19.5 g of F CH2 COOH (molar mass = 78 g mol–1) is dissolved in

500 g of water, the depression in freezing point is observed to be 1°C.
Calculate the degree of dissociation of F CH2 COOH. [Given : Kf for water =
1.86 K kg mol–1] (3/5, 2020)

91. The freezing point of a solution containing 5 g of benzoic acid (M = 122 g

mol–1) in 35 g of benzene is depressed by 2.94 K. What is the percentage
association of benzoic acid if it forms a dimer in solution? (Kf for benzene =
4.9 K kg mol–1) (2020)

92. Calculate the freezing point of solution when 1.9 g of MgCl2(M = 95 g

mol–1) was dissolved in 50 g of water, assuming MgCl2 undergoes complete
ionization. (Kf for water = 1.86 K kg mol–1) (Delhi 2016)
 93. When 2.56 g of sulphur was dissolved in 100 g of CS2, the freezing
point lowered by 0.383 K. Calculate the formula of sulphur (Sx). (Kf the CS2
= 3.83 K kg mol–1, atomic mass of sulphur = 32 g mol–1) (3/5 Delhi 2016)

94. Calculate the boiling point of solution when 4 g of MgSO4 (M = 120 g

mol–1) was dissolved in 100 g of water, assuming MgSO4 undergoes
complete ionization. (Kb for water = 0.52 K kg mol–1) (AI 2016)

95. 3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in

freezing point of 1.62 K. Calculate the van’t Hoff factor and predict the
nature of solute (associated or dissociated). (Given : Molar mass of benzoic
acid = 122 g mol–1, Kf for benzene = 4.9 K kg mol–1) (Delhi 2015)

96. Calculate the mass of NaCl (molar mass = 58.5 g mol–1) to be dissolved
in 37.2 g of water to lower the freezing point by 2°C, assuming that NaCl
undergoes complete dissociation. (Kf for water = 1.86 K kg mol–1) (Foreign
2015)

97. Determine the osmotic pressure of a solution prepared by dissolving 2.5

× 10–2 g of K2SO4 in 2 L of water at 25°C, assuming that it is completely
dissociated. (R = 0.0821 L atm K–1 mol–1, molar mass of K2SO4 = 174 g
mol–1) (Delhi 2013)

98. Calculate the amount of KCl which must be added to 1 kg of water so

that the freezing point is depressed by 2 K (the Kf for water = 1.86 K kg
mol–1). (Delhi 2012)

99. 15.0 g of an unknown molecular material was dissolved in 450 g of

water. The resulting solution was found to freeze at –0.34°C. What is the
molar mass of this material? (Kf for water = 1.86 K kg mol–1) (Delhi 2012,
3/5, AI 2012)

100. Calculate the freezing point of an aqueous solution containing 10.50 g

of MgBr2 in 200 g of water. (Molar mass of MgBr2 = 184 g mol–1, Kf for
water = 1.86 K kg mol–1) (3/5, Delhi 2011)

101. A 0.561 m solution of an unknown electrolyte depresses the freezing

point of water by 2.93°C. What is van’t Hoff factor for this electrolyte? The
freezing point depression constant (Kf) for water is 1.86°C kg mol–1.
(Foreign 2011)

102. Phenol associates in benzene to a certain extent to form a dimer. A

solution containing 20 g of phenol in 1.0 kg of benzene has its freezing point
lowered by 0.69 K. Calculate the fraction of phenol that has dimerised [Given
Kf for benzene = 5.1 K m–1] (3/5, Delhi 2011C)

103. An aqueous solution containing 12.48 g of barium chloride in 1.0 kg of

water boils at 373.0832 K. Calculate the degree of dissociation of barium
chloride. [Given Kb for H2O = 0.52 K m–1; Molar mass of BaCl2 = 208.34 g
mol–1] (3/5, Delhi 2011C)

104. What mass of NaCl must be dissolved in 65.0 g of water to lower the
freezing point of water by 7.50°C? The freezing point depression constant
(Kf) for water is 1.86°C/m. Assume van’t Hoff factor for NaCl is 1.87. (Molar
mass of NaCl = 58.5 g mol–1) (AI 2011)

Answer:
 31. Azeotropes are the
binary mixtures of solutions
that have the same
composition in liquid and
vapour phases and that
have constant boiling point.

32. Refer to answer 31.

33. The differences between

ideal solutions and non
ideal solutions are as
follows : (i) In ideal solutions
∆Vmixing = 0 and ∆Hmix=
0 whereas in non ideal
solutions, ∆Vmix ≠ 0 and
∆Hmix ≠ 0. (ii) In ideal
solutions, each component
obeys Raoult’s law at all
temperatures and
concentrations whereas in
non ideal solutions, they do
not obey Raoult’s law.

34. Mixture of chloroform

and acetone shows
negative deviation from
Raoult’s law, thus it forms
maximum boiling
azeotrope. This is because
chloroform molecule is able
29. A solution which obeys Raoult’s law of
to form hydrogen bond with
vapour pressure for all compositions is called
acetone molecule
ideal solution. In this solution ∆Vmix = 0, ∆Hmix
= 0 A .... B interaction = A .... A and B .... B
interactions.

30. Non-ideal solutions that show negative

deviation from Raoult’s law form maximum boiling azeotropes
35. Positive deviation : For non-ideal solutions, if the vapour pressure is
higher, then it is said to exhibit positive deviation. AB interactions are
weaker than AA or B B interactions. Due to this, vapour pressure increases
which results in positive deviation. In positive deviation, intermolecular force
decreases, volume increases, vapour pressures increases. enthalpy
increases. Therefore, ∆Hmix = +ve, ∆Vmix = + ve. e.g., ethanol + acetone
and carbon disulphide + acetone show positive deviation. (Graph)

36. Refer to answer 31. A minimum boiling azeotrope is formed by solutions

showing a large positive deviation from Raoult’s law at a specific
composition. For example an ethanol-water mixture containing
approximately 95% ethanol by volume.

37. Negative deviation : For non-ideal solution, if the vapour pressure is

lower, then it is said to exhibit negative deviation. AB interactions are
stronger than AA and B B interactions. Due to this, vapour pressure
decreases which results in negative deviation. In negative deviation,
intermolecular force increases, volume decreases, vapour pressure
decreases and heat is released. Therefore, ∆Hmix = –ve, ∆Vmix = – ve. For
example, phenol + aniline and chloroform + acetone show negative
deviation. (graph)

38. Refer to answer 31. A maximum boiling azeotrope is formed by solutions

showing a large negative deviation from Raoult’s law at a specific
composition. For example, chloroform + acetone mixture.

39. Refer to answer 28

40. Refer to answers 35 and 37.

41. Refer to answers 34.

42. Colligative properties of ideal solutions depend only on the number of

particles of solute dissolved in a definite amount of the solvent and do not
depend on the nature of solute.

43. When the external pressure applied becomes more than the osmotic
pressure of solution then the solvent molecules from the solution pass
through the semi-permeable membrane to the solvent side and the process
is called reverse osmosis.

44. In osmotic pressure method, pressure can be measured at room

temperature and the molarity of the solution is used instead of molality. That
is why this method is used for determination of molar masses of
macromolecules as they are generally not stable at higher temperatures.

45. Properties which depend upon the number of solute particles irrespective
of their nature relative to the total number of particles present in the solution
are called colligative properties.

46. Osmotic pressure is the extra pressure which is applied on the solution to
just prevent the flow of solvent into the solution through a semipermeable
membrane.

47. Two solutions having same osmotic pressure at a given temperature are
called isotonic solutions.

48. Molal elevation constant may be defined as the elevation in boiling point
when the molality of the solution is unity (i.e., 1 mole of the solute is
dissolved in 1 kg (1000 g) of the solvent). The units of Kb are therefore,
degree/molality i.e., K/m or °C/m or K kg mol–1.

49. When a non-volatile solute is added to a solvent, the vapour pressure of

the solvent (above the resulting solution) is lower than the vapour pressure
above the pure solvent.
53. (i) As KCl is an electrolyte and
50. Refer to answer 48.
one formula unit of KCl dissociates
to give two ions (K+ and Cl–),
therefore molar concentration of
particles in the solution = 0.1 × 2 M
= 0.2 M As elevation of boiling point
(or any colligative property) is
directly proportional to number of
particles in solution, hence 0.1 M
KCl has higher boiling point that 0.1
M glucose. (ii) Salting is used
because most bacteria, fungi and
other potentially pathogenic
organism cannot survive in a highly
salty environment, due to the
hypertonic nature of salt. Any living
cell in such an environment will
become dehydrated through
osmosis and die or become
temporarily deactivated.
55. (i) The elevation in boiling point of a solution is a colligative property
which depends on the number of moles of solute added. Higher the
concentration of solute added, higher will be the elevation in boiling point.
Thus, 2 M glucose has higher boiling point than 1 M glucose solution. (ii)
Refer to answer 43.

56. (i) 1.2% sodium chloride solution is hypertonic with respect to 0.9%
sodium chloride solution or blood cells thus, on placing blood cells in this
solution exosmosis takes place that results in shrinking of cells. (ii) 0.4%
sodium chloride solution is hypotonic with respect to 0.9% sodium chloride
solution or blood cells thus, on placing blood cells in this solution endosmosis
takes place that results in swelling of cells.

57. The boiling point of the solution is always higher than that of the pure
solvent. As the vapour pressure of the solution is lower than that of the pure
solvent and vapour pressure increases with increase in temperature, hence,
the solution has to be heated more to make the vapour pressure equal to the
atmospheric pressure. Elevation of boiling point is a colligative property
because it depends on number of solute particles present in a solution
60. When a non-volatile solute is added to a solvent, the freezing point of the
solution is always lower than that of pure solvent as the vapour pressure of
the solvent decreases in the presence of non-volatile solute. Plot for the
lowering in freezing point of water when NaCl is added to it is shown as :
(Depression of freezing point graph)

61. Osmosis : The spontaneous movement of the solvent molecules from the
pure solvent or from a dilute solution to a concentrated solution through a
semi-permeable membrane is called osmosis. Osmotic Pressure : The
minimum excess pressure that has to be applied on the solution to prevent
the passage of solvent molecules into it through semipermeable membrane
is called osmotic pressure. Osmotic pressure is a colligative property
because it depends on the number of solute particles and not on their
nature.

62. (i) Number of particles of solute (ii) Association or dissociation of solute

(iii) Concentration of solute (iv) Temperature
71. (i) Refer to answer 46. (ii) Refer to answer 45.