CHELLAMMAL VIDYALAYA

(A unit of Sowdambikaa Group of Schools)

Thiruverumbur – Kallanai Road, Vengur, Trichy – 620013
Affiliated to the Central Board of Secondary Education, New Delhi

CHEMISTRY CHAPTER 1- SOLUTION QUESTIONBANK

1. (i)On mixing liquid X and liquid Y, the volume of the resulting solution
increases. What type of deviation from Raoult’s law is shown by the resulting
solution? What change in temperature would you observe after mixing liquids X
and Y?
(ii) How can the direction of osmosis be reversed? Write one use of reverse
osmosis.
Answer: (i) Positive deviation. The temperature will decrease.
(ii) When we apply pressure on solution side more than osmotic pressure, osmosis can
be reversed. It is used for desalination of sea water.

2. What do you understand by depression of freezing point? Derive the relationship

between depression of freezing point and molar mass of the solute.
Answer: The decrease in freezing point when non-volatile solute is added to a solvent, is
called depression in freezing point.

The pressure exerted by an individual gas

in a mixture is known as its partial
pressure formula.

3. State Henry’s law. Why do gases

always tend to be less soluble in liquids
as
the temperature is raised?
Answer: Henry’s law states, ‘The partial pressure of the gas dissolved in a liquid is
directly proportional to its mole fraction. pgas= KH xgas, where xgas is the mole fraction of
gas and pgas is the partial pressure of the gas.
When temperature is increased, KH (Henry’s law constant) increases, therefore, the
solubility of gases in liquid decreases.

4. State Raoult’s law for the solution containing volatile components. Write two
differences between an ideal solution and a non-ideal solution.
Answer: Raoult’s law: It states that partial vapour pressure of each component is
directly proportional to their mole fraction if both solute and solvent are volatile.

5. 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?
Answer: When we add a non-volatile solute to a pure solvent, the vapour pressure of
solution decreases, therefore, it is to be heated to higher temperature so that its vapour
pressure becomes equal to the atmospheric pressure, i.e. its boiling point will be higher.
Elevation of boiling point is a colligative property because it depends upon the number of
particles of solute and not on nature of solute.

6. Derive the relationship between relative lowering of vapour pressure and molar
mass of the solute.
Answer:

7. What is meant by positive deviations from Raoult’s law? Give an example. What
is the sign of ΔmixH for positive deviation?
Answer: Those solutions in which force of attraction between A—B is less than A—A and
B—B, shows positive deviation from Raoult’s law, e.g. ethanol and water show positive
deviation from Raoult’s law.

. Define azeotropes. What type of azeotrope is formed by positive deviation from

Raoult’s law? Give an example.
Answer: Azeotropes are the constant boiling mixtures which distill out unchanged in
their composition.
Minimum boiling azeotropes are formed by the solutions showing positive deviation, e.g.
cyclohexane and ethanol.

9. (i) On mixing liquid X and liquid Y, volume of the resulting solution decreases.
What type of deviation from Raoult’s law is shown by the resulting solution? What
change in temperature would you observe after mixing liquids X and Y? .
(ii) What happens when we place the blood cell in water (hypotonic solution)? Give
reason.
Answer: (i) The resulting solution will show negative deviation from Raoult’s law. The
temperature of solution will increase.
(ii) The cell will swell and even may burst due to inflow of solvent because osmosis.

10. What is meant by negative deviation from Raoult’s law? Give an example. What
is the sign of AmixH for negative deviation?
Answer: A solution is said to deviate negatively from Raoult’s law if the forces of
attraction between A—B are more than A—A and B—B, e.g. CHCl3and CH3COCH3.

11. Define azeotropes. What type of azeotrope is formed by negative deviation

from Raoult’s law. Give an example.
Answer: Azeotropes are the constant boiling mixtures which distill out unchanged in
their composition.
Maximum boiling azeotropes are formed by the solution showing negative deviation, e.g.
H2O and HCl.

12. Vapour pressure of

water at 20 °C is 17.5
mm Hg. Calculate the
vapour pressure of
water at 20 °C when 15
g of glucose (Molar
mass = 180 g mol-1) is
dissolved in 150 g of
water.
Answer:

13. A solution is prepared by dissolving 5 g of non-volatile solute in 95 g of water.

It has a vapour pressure of
23.375 mm Hg at 25°C.
Calculate the piolar mass of
the solute, (vapour pressure of
pure water at 25 °C is 23.75
mm Hg).
Answer:
14. When 1.5 g of a non-volatile
solute was dissolved in 90 g of
benzene, the boiling point of
benzene raised from 353.23 K to
353.93 K. Calculate the molar
mass of the solute. (Kb for
benzene = 2.52 kg mol-1)
Answer:

15. Calculate the freezing point of the solution when 31 g of ethylene glycol
(C2H6O2) is dissolved in 500 g of water. (Ky for water = 1.86 K kg mol-1)
Answer:

16. 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, Ky for
benzene = 4.9 K kg mol-1)
Answer:

17. A solution is prepared by dissolving 10 g of non-volatile solute in 200 g of

water. It has a vapour pressure of 31.84 mm Hg at 308 K. Calculate the molar mass
of the solute. (Vapour pressure of pure water at 308 K = 32 mm Hg)
Answer:

18. 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. (Ky for water = 1.86 K kg mol-1)
Answer:

19. What possible value of ‘i’ will it have if solute molecules undergo association
in solution?
Answer: i < 1, if solute molecules
undergo association.

20. Some liquids on mixing form

‘azeotropes’. What are ‘azeotropes’?
Answer: Azeotropes are the constant
boiling mixtures which distill out
unchanged in their composition.

21. What type of intermolecular attractive interaction exists in the pair of

methanol
and acetone?
Answer: H-bonding exists between methanol and acetone. It also has dipole-dipole
attraction.

22. What are isotonic solutions?

Answer: Those solutions which exert same osmotic pressure because they have same
molar concentrations and same number of particles are called isotonic solutions.

23. How is vapour pressure of solvent affected when a non-volatile solute is

dissolved in it?
Answer: Vapour pressure of solution decreases because surface consists of bpth solute
and solvent molecules. The escaping tendency of solvent into vapours decreases.

24. Differentiate between molarity and molality of solution. How can we change
molality value into molarity value.
Answer: Molarity is the number of moles of solute
dissolved per litre of solution, whereas molality is the
number of moles of solute dissolved per kg of solvent.

25. 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
(Kj = 5.12 K kg mol-1).
Answer:

26. Define an ideal solution and write one of its characteristics.

Answer: Ideal solution is a solution which follows Raoult’s law.
Characteristics:

27. State Henry’s law. What is the effect of temperature on the solubility of a gas
in a liquid?
Answer: Henry’s Law: The mole fraction of a gas in a solution (solubility of gas in liquid)
is directly proportional to partial pressure of gas over solution.
where KH is Henry’s law constant, %gas is mole fraction of gas
and ^gas is the partial pressure of gas.
The solubility of a gas decreases with increase in temperature.

28. State Raoult’s law for the solution containing volatile components. What is the
similarity between Raoult’s law and Henry’s law?

Answer:
29. 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°C, what is
the boiling point
of the solution.
[Kft = 0.52 K kg
mol-1 and Ky =
1.86 K kg mol-
1 for H2O.

30. (a) Define the following terms:

(i) Molarity (ii) Molal
elevation constant
(Kb)
(b) 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.
31. (a) What type of
deviation is shown
by a mixture of
ethanol and
acetone? Give
reason.
(b) 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 mL1 )

32. (a) State Raoult’s

law for a solution
containing volatile
components. Name the
solution which follows
Raoult’s law at all
concentrations and
temperatures.
(b) Calculate the boiling
point elevation for a
solution prepared by
adding 10 g of CaCl2 to
200 g of water. (Kb for
water = 0.512 K kg mol-
1, Molar mass of CaCl2 = 111 g mol-1)

33.(a) Define the following terms:

(i) Azeotrope (ii) Osmotic pressure (iii) Colligative properties
(b) Calculate the molarity of 9.8% (w/w) solution of H2S04 if the density of the
olution is 1.02 g mL-1 (Molar mass of H2S04 = 98 g mol-1).Answer:

34. Define osmotic pressure.

Answer: Refer Ans. to Q.33 (ii).

35. Measurement of which colligative property is preferred for determination of

molar mass of biomolecules?
Answer: Osmotic pressure is preferred for determining molecular weight of biomolecules.

36. What type of deviation is shown by a mixture of ethanol and acetone? What
type of azeotrope is formed by mixing ethanol and acetone?
Answer: Positive deviation is shown by a mixture of ethanol and acetone. Minimum
boiling azeotropes are formed.

37. Out of two 0.1 molal solutions of glucose and of potassium chloride, which one
will have a higher boiling point and why?
Answer: 0.1 molal KC1 solution will have higher boiling point, because KC1 dissociates
into K+ and Cl–ions. The number of particles are double than 0.1 m glucose solution,
therefore, higher ATa, hence higher the boiling point.

38. 18 g of glucose,
C6H1206 (Molar Mass = 180
g mol-1) is dissolved in 1 kg
of water in a sauce pan. At
what temperature will this
solution boil? (Kh for water
= 0.52 K kg mol-1, boiling
point of pure water =
373.15 K)
Answer:
39. Henry’s law constant
(KH) for the solution of
methane in benzene at
298 K is 4.27 X 105
mmHg. Calculate the
solubility of methane in
benzene at 298 K under
760 mm Hg.
Answer:

40. (i) Why is an increase

in temperature observed
on mixing chloroform and
acetone?
(ii) Why does sodium chloride solution freeze at a lower temperature than water?
Answer: (i) It is because force of attraction between chloroform and acetone is more than
the force of attraction between CHC13—CHC13 or Acetone-Acetone. Therefore, AH = -ve,
i.e. exothermic process, therefore, there is increase in temperature.
(ii) It is because when NaCl is added, vapour pressure of solution becomes less and at a
lower temperature, vapour pressure of solid and solution will become equal, i.e. freezing
point is lowered.

41. Determine the osmotic pressure of a solution prepared by dissolving 2.5 X 10-
2 g of K2S04 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).

Answer:

42. (a) State Raoult’s law for a solution containing volatile components. How does
Raoult’s law become a special case of Henry’s law?
(b) 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the
freezing point of benzene by 0.40 K. Find the molar mass of the solute. (Kyfor
benzene = 5.12 K kg mol-1)
Answer:

43. (a) Define the following terms: ..

(i) Ideal solution (ii) Azeotrope (iii) Osmotic pressure
(b) A solution of glucose (C6H1206) in water is labelled as 10% by weight. What
would be the molality of the solution?
(Molar mass of glucose = 180 g mol-1)
Answer:

44. Define Ebullioscopic constant.

Answer: Ebullioscopic Constant (Boiling Point Elevation Constant): It is equal to the
elevation in boiling point of 1 molal solution, i.e. 1 mole of solute is dissolved in 1 kg of
solvent. The unit of is K molal-1 or KJm or K kg mol-1

45. State Raoult’s law for a solution containing volatile components.

Answer: The vapour pressure of each component is direcdy proportional to mole fraction
of each component.

46. Define ‘mole fraction’ of a substance in a solution.

Answer: Mole fraction of a substance is defined as the ratio of number of moles of
substance to the total number of moles of solute and solvent, i.e. all the substances.

47. Why are aquatic species more comfortable in cold water in comparison to warm
water?
Answer: Cold water has more dissolved oxygen than warm water because solubility of
gas in liquid decreases with increase in temperature.
48. 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. (Kft for water = 0.512 K
kg mol-1)
Answer:

49. Define the following terms:

(i)Mole fraction (ii)Isotonic solutions (iii)van’t Hoff factor (iv)Ideal solution
Answer:

50. The density of water of a lake is 1.25 g mL-1 and one kg of this water contains
92 g of Na+ ions. What is the molarity of Na+ ions in the water of the lake? (Atomic
mass of Na = 23.00 u)
Answer:

51. Calculate the amount of KC1

which must be added to 1 kg of
water so that the freezing point is
depressed by 2 K. (Ky for water =
1.86 kg mol-1)

52. At 25 °C the
saturated vapour
pressure of water is
3.165 kPa (23.75 mm
Hg). Find the
saturated vapour
pressure of a 5%
aqueous solution of
urea (carbamide) at
the same
temperature. (Molar
mass of urea = 60.05
g mol-1)
Answer:

53. Calculate the freezing point

depression expected for 0.0711m
aqueous solution of Na2S04. If
this solution actually freezes at –
0.320 °C, what would be the
value of van’t Hoff factor? (Ky for
water is 1.86 °C mol-1).
Answer:

54. (a) Explain the following:

(i) Henry’s law about dissolution of a gas in a liquid.
{ii) Boiling point elevation constant for a
solvent.
(b) A solution of glycerol (C3Hg03) in water
was prepared by dissolving some glycerol
in 500 g of water. This solution has a
boiling point of 100.42 °C. What mass of
glycerol was dissolved to make this
solution? (Kf for water = 0.512 K kg mol-1)
Answer: (a) (i) Henry’s Law: It states that the
solubility of a gas in a liquid is directly
proportional to the pressure of the gas.
p = KHx, where p is pressure of gas, x is the
mole fraction of the gas and KH is Henry’s law
constant.
{ii) Boiling Point Elevation Constant (Molal Boiling Point Elevation Constant). It is equal
to the elevation in boiling point of 1 molal solution, i.e. 1 mole of solute is dissolved in 1
kg of solvent. It is also called ebullioscopic constant. The unit of is K/m or K kg mol-1.
55. State Raoult’s law in its general form with respect to solutions.
Answer: The partial vapour pressure of each component in solution is directly
proportional to its mole fraction if both components are volatile.

56. Define Henry’s law about solubility of a gas in a liquid.

Answer: Henry’s law states that partial pressure of the gas is directly proportional to its
mole fraction.

57. What is meant by ‘reverse osmosis’?

Answer: Reverse Osmosis: If extra pressure is applied on the solution side and exceeds
the osmotic pressure, the osmosis can be reversed. That is, pure water can be forced out
of the solution to pass through the pores of the membrane in the opposite direction. This
is called reverse osmosis.

58. State Raoult’s law for a solution containing volatile components.

Answer: Refer Ans. to Q. 45.

59. 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

Answer:

60.What would be the molar mass of a compound if 6.21 g of it dissolved in 24.0 g

of chloroform forms 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, for
chloroform is 3.63
°C/m.

61. What mass of NaCl (molar mass = 58.5 g mol-1) must be dissolved in 65 g of
water to lower the freezing point by 7.50 °C? The freezing point depression
constant, Kffor water is 1.86 K kg mol-1. Assume van’t Hoff factor for NaCl is 1.87.
Answer:

62. (a) Differentiate between molarity and molality for a solution. How does a
change in temperature influence their values?
(b) 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 is 1.86 K
kg mol-1)
Answer:

63. (a) Define the terms osmosis and osmotic pressure. Is osmotic pressure of a
solution a colligative property? Explain.
(b) Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to
250.0 g of water.
(Kf for water = 0.512 K kg mol-1, Molar mass of NaCl = 58.44 g)
Answer: (a) Osmosis: When a solvent is separated from the solution by a semi-
permeable
membrane which
allows the passage
of solvent molecules
but does not allow
solute particles to
pass through it,
there is net flow of
solvent molecules from the.solvent to the solution which is called osmosis.
Osmotic Pressure: Osmotic pressure may be defined as extra pressure that must be
applied to the solution side to prevent osmosis (the flow of solvent

into solution through a semi-permeable membrane.)

64. (a) State the following:
(i) Henry’s law about partial pressure of a gas in a mixture.
(ii) Raoult’s law in its general form in reference to solutions.
(b) 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 non-electrolyte, determine its molar mass.
Answer: (i) Henry’s Law. It states that the partial vapour
pressure of a gas in vapour phase is directly proportional to the
mole fraction of the gas in the solution.
where KH is Henry’s law constant,
x is mole fraction of gas in solution
and p is partial vapour pressure of
the gas in solution.
(ii) Raoult’s Law for solution of
Non-volatile Solute. The relative
lowering of vapour pressure for a
solution is equal to the mole
fraction of solute when solvent
alone is volatile.

65. (a) The molecular masses of polymers are determined by osmotic pressure
method and not by measuring other colligative properties. Give two reasons.
(b) At 300 K, 36 g of glucose, C6H1206 present per litre in its solution has an
osmotic pressure of 4.98 bar. If the osmotic pressure of another glucose solution is
1.52 bar at the same temperature,
calculate the concentration of the
other solution.
Answer: (a) (i) It is measured at room
temperature. (ii)
It has appreciable value.

66. (a) List any four factors on which the colligative properties of a solution
depend.
(b) Calculate the boiling point of one molar aqueous solution (density 1.06 gmL-1)
of KBr. [Given: Kb for H2O = 0.52 K kg mol-1, Atomic mass: K = 39, Br = 80]
Answer:

67. (a) Explain why a solution of chloroform and acetone shows negative deviation
from Raoult’s law.
(b) Phenol associates in benzene to 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 Km-1]
Answer: (a) It is due to
formation of H-bonding due
to which escaping tendency
of molecules and vapour
pressure of the solution
decrease, therefore, boiling
point of the solution
increases. Hence, such
solution shows negative
deviation from Raoult’s law.
(b) ATy = 0.69 K, Ky = 5.1
K/m,
WB = 20 g, WA = 1 kg =
1000 g

68. (a) What is van’t Hoff factor? What possible values can it have if the solute
molecules undergo dissociation?
(b) 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
for H20 = 0.52 K m-1; Molar mass of BaCl2= 208.34 g mol-1]
Answer:
69. How is the molality of a solution different from its molarity?
Answer: Molality is defined as the number of moles ofsolute dissolved per kg ofsolvent,
whereas
molarity is defined as the number of moles of solute dissolved in per litre of solution.

70. State Raoult’s law for a solution containing volatile components.

Answer: Refer Ans. to Q. 45.

71. Non-ideal solutions exhibit either positive or negative deviations from Raoult’s
law. What are these deviations and how are they caused?
Answer: Positive Deviation from Raoult’s Law: In those non-ideal solutions, in which
partial pressure of each component A and B is higher than that calculated from Raoult’s
1? show positive deviation from Raoult’s law, e.g. water and ethanol. Negative Deviation
from Raoult’s Lawr: The partial vapour pressure of component A is found to be less than
calculated from Raoult’s law’ on adding the second component B and when A is added to
B, the partial vapour pressure of component B is found to be less than that calculated
from Raoult’s law.

72. State Henry’s law and mention its two important applications.
Answer: Henry’s Law: It states that the partial vapour pressure of gas in vapour phase is
directly proportional to the mole fraction of the gas in the solution.
Applications of Henry’s Law:
(i) To minimise the painful effects of deep sea divers, oxygen diluted with less soluble
helium gas is used as breathing gas instead of air because nitrogen is more soluble in
blood and cause pain or bends.
(ii) To increase the solubility of COs in soft drinks and soda water, the bottle is sealed
under high pressure.

73. Find the boiling point of a

solution containing 0.520 g of
glucose (C6H1206) dissolved in
80.2 g of water. [Given: Kf for
water = 0.52 K/m]
Answer:

74. Define the term, ‘molarity of a solution’. State one disadvantage in using the
molarity as the unit of concentration.
Answer: Molarity is defined as the number of moles of solute dissolved per litre of
solution. Its disadvantage is that it changes with change in temperature and volume of
solute is not taken into account.

75. What mass of ethylene

glycol (molar mass = 62.0 g
mol-1) must be added to 5.50
kg of water to lower the
freezing point of water from 0
°C to -10.0 °C? (Kf for water =
1.86 K kg mol-1)
Answer:

76. 15 g of an unknown molecular substance was dissolved in 450 g of water. The

resulting solution freezes at – 0.34 °C. What is the molar mass of the substance?
(Kf for water = 1.86 K kg mol-1)
Answer:

77. (a) Define the terms osmosis and osmotic pressure. What is the advantage of
using osmotic pressure as compared to other colligative properties for the
determination of molar masses of solutes in solutions?
(b) A solution prepared from 1.25 g of oil of wintergreen (methyl salicylate) in 90.0
g of benzene has a boiling point of 80.31 °C. Determine the molar mass of this
compound. (Boiling point of pure benzene = 80.10 °C and K& for benzene = 2.53 °C
kg mol-1)
Answer: (a) Osmosis: When a solution is separated from the solvent by a semi-
permeable membrane which allows the passage of solvent molecules but does not allow
solute particles to pass through it, there is net flow of solvent molecules from the solvent
to the solution which is called osmosis.
Osmotic Pressure: Osmotic pressure may be defined as extra pressure that must be
applied to the solution
side to prevent the flow
of solvent into solution
through a semi-
permeable membrane.
Osmotic pressure is
determined at room
temperature and has
appreciable value which
can be easily measured.

78. State ‘Raoult’s Law’ for a solution of volatile liquids.

Answer: Refer Ans. to Q.45.

79. Define an ideal solution.

Answer: Refer Ans. to Q.43 (a) (i).

80. Measurement of which colligative property is preferred for determination of

molar mass of biomolecules?
Answer: Refer Ans. to Q.35.

81. What is meant by‘reverse osmosis’?

Answer: Refer Ans. to Q.57.

82. 100 mg of a protein is dissolved

in just enough water to make 10.0
mL of solution. If this solution has
an osmotic pressure of 13.3 mm Hg
at 25 °C, what is the molar mass of
the protein? (R = 0.0821 L atm
mol-1K-1and 760 mm Hg = 1 atm)
Answer:

83. Calculate the freezing point depression expected for 0.0711 m aqueous
solution of Na2S04. If this solution actually freezes at – 0.320 °C, what would be the
value of van’t Hoff factor? (Ky for water is 1.86 °C mol-1).
Answer: Refer Ans. to Q.53.

84. Calculate the amount of sodium

chloride which must be added to one
kilogram of water so that the
freezing point of water is depressed
by 3 K. [Given: Kf = 1.86 K kg mol-1,
Atomic mass: Na = 23.0, Cl = 35.5]
85. A solution of urea in water
has a boiling point of 373.128 K.
Calculate the freezing point of
the same solution. [Given: For
water, Kf = 1.86 Km-1,Kb = 0.52
Km-1]
Answer:

86. 0.1 mole of acetic acid was dissolved in 1 kg of benzene. Depression in freezing
point of benzene was determined to be 0.256 K. What conclusion can you draw
about the state of the solute in solution?
Answer:
87. (a) What is meant by:
(i) Colligative properties (ii) Molality of a solution
(b) What concentration of nitrogen should be present in a glass of water at room
temperature? Assume a temperature of 25 °C, a total pressure of 1 atmosphere and
mole fraction of nitrogen in air of 0.78. [KHfor nitrogen = 8.42 X 10-7M/mm Hg]
Answer: (a) (i) Colligative properties: Those properties of solutions which depend upon
the number of particles of solute and solvent but not on the nature of solute are called
colligative properties.
(ii) Molality of a solution: It is
defined as the number of
moles of solute dissolved in
per 1000 g or 1 kg of solvent.