# SOLUTIONS – 6

1) The normality of 2 M sulphuric acid is
a) 2 N
b) 4 N
c) N/2
d) N/4

2) How many grams of a dibasic acid (mol. wt.=200) should be present in 100 ml of its aqueous solution to give decinormal strength?
a) 1 g
b) 2 g
c) 10 g
d) 20 g

3) A 500 g toothpaste sample has 0.2 g fluoride concentration. What is the concentration of fluorine in terms of ppm level?
a) 250
b) 200
c) 400
d) 1000

4) An aqueous solution of glucose is 10% in strength. The volume in which one gram mole of it is dissolved will be
a) 18 litres
b) 9 litres
c) 0.9 litrs
d) 1.8 litres

5) The molality of a sulphuric acid solution in which the mole fraction of water is 0.85 is
a) 9.80
b) 10.50
c) 10.58
d) 11.25

6) The concentration of an aqueous solution of 0.01 M solution is very nearly equal to which of the following?
a)
b)
c)
d)

7) Which amongst the following has the highest normality?
a) 8.0 g of KOH/100 ml of solution
b)
c)
d) 6 g of NaOH/100 g of water

8) The number of iodine atoms present in of its 0.1 M solution is
a)
b)
c)
d)

9) A molal solution of sodium chloride has a density of 1.21 g . The molarity of this solution is
a) 4.15
b) 1.143
c) 2.95
d) 3.15

10) The normality of 10% (weight/volume) acetic acid is
a) 1 N
b) 10 N
c) 1.7 N
d) 0.83 N

11) 1000g solution contains 10 g carbonate. The concentration of solution is
a) 10000ppm
b) 1000 ppm
c) 100 ppm
d) 10 ppm

12) 10 ml. concentrated sulphuric acid (18 molar) is diluted to 10 litres. The approximate strength of the acid would be
a) 0.18 N
b) 0.036 N
c) 0.36 N
d) 0.09 N

13) 100 ml of 0.3 N HCI is mixed with 200 ml of 0.6 N . The final normality of the resulting solution will be
a) 0.1 N
b) 0.2 N
c) 0.3 N
d) 0.5 N

14) To 4 L of 0.2 M solution of NaOH, 2 L of 0.5 M NaOH are added. The molarity of resulting solution is
a) 0.9 M
b) 0.3 M
c) 1.8 M
d) 0.18 M

15) 5.85 g of NaCl is dissolved in and solution made upto 500 ml. The molarity is
a) 0.1
b) 0.2
c) 1
d) 0.117

16) For preparing 0.1 N solution of a compound from its impure sample of which the percentage purity is known, the weight of the substance required will be
a) more than the theoretical weight
b) less than the theoretical weight
c) same as the theoretical weight
d) none of the above

17) What is the molarity of solution that has a density 1.84 gm/cc at and contains 98% by weight?
a) 4.18 M
b) 8.14 M
c) 18.4 M
d) 18 M

18) An X molal solution of a compound in benzene has mole fraction of solute equal to 0.2. The value of X is
a) 14
b) 3.2
c) 1.4
d) 2.0

19) The amount of oxalic acid (molecular weight: 63) required to prepare 500 ml of its 0.10 N solution is
a) 0.315 g
b) 3.150 g
c) 6.300 g
d) 63.00 g

20) The molarity of an aqueous solution of NaOH containing 8 g in 2 litre of the solution is
a) 0.1 M
b) 0.2 M
c) 0.25 M
d) 0.15 M

21) An aqueous solution of urea containing 18 g urea in 1500 of the solution has a density equal to 1.052. If the molecular weight of urea is 60, then the molality of the solution is
a) 0.200
b) 0.192
c) 0.100
d) 1.200

22) 12 g was dissolved in water. The volume of the solution was made 1200 ml. The normality of the solution is
a) 0.05 N
b) 0.102 N
c) 0.112 N
d) 0.204 N

23) The normality of 1 is
a) 1 N
b) 0.5 N
c) 2 N
d) 3 N

24) An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 ml. The volume of 0.1 N NaOH required to completely neutralize 10 ml of this solution is
a) 40 ml
b) 20 ml
c) 60 ml
d) 4 ml

25) 98 g is present in one litre. The solution is
a) 1 molar
b) 1 normal
c) 1 molal
d) 2 molar

26) 3.0 molal sodium hydroxide solution has a density of . The molarity of this solution is
a) 2.9732
b) 3.0504
c) 3.6485
d) 4.110

27) Rectified spirit contains 95% ethanol by weight. The mole fraction of ethanol will be
a) 0.881
b) 0.99
c) 0.118
d) 0.81

28) 10.6 g of are present in 188.8 g of the aqueous solution. The mole fraction of sodium carbonate will be
a) 0.0100
b) 0.0094
c) 0.0101
d) 0.0091

29) The volume of 0.2 M solution containing 20 millimoles of the solute is
a)
b)
c)
d)

30) To 5.85 g of NaCl, one kg of water is added to prepare a solution. What is the strength of NaCl in this solution? (Molecular weight of NaCl = 58.5)
a) 0.1 normal
b) 0.1 molal
c) 0.1 molar
d) 0.1 formal

31) 100 ml of 0.1 M solution of solute A are mixed with 200 ml of 0.1 M solution of solute B. If A and B are non-reacting substances, the molarity of the final solution will be
a) 0.3 M
b) 0.4 M
c) 0.1 M
d) 0.15 M

32) Concentrated sulphuric acid is approximately 18 molar: 5 cc of it are added to make 500 cc of the solution. The approx. normality of the solution will be
a) 0.18
b) 0.09
c) 0.36
d) 0.27

33) The molarity of a solution prepared by adding 7.1 g of (formula weight 142 amu) to enough water to make 100 ml volume is
a) 2.0 M
b) 1.0 M
c) 0.5 M
d) 0.05 m

34) The mole fraction of glycerine in a solution containing 36 g water and 46 g of glycerine is (working)
a) 0.46
b) 0.40
c) 0.20
d) 0.36

35) The number of milli equivalents in 60 ml 4.0 M is
a) 240
b) 480
c) 24
d) 48

36) The mole fraction of (molecular weight = 46) in 5 molal aqueous ethyl alcohol solution is
a) 0.082
b) 0.82
c) 5
d) 5/55.55

37) The molarity of pure water is
a) 55.6
b) 50
c) 100
d) 18

38) If 18 g of glucose is present in 1018 g of an aqueous solution of glucose, it is said to be
a) 1 molal
b) 1.1 molal
c) 0.5 molal
d) 0.1 molal

39) 0.115 g of pure sodium metal was dissolved in 500 ml distilled water. The normality of the above solution would be
a) 0.010 N
b) 0.0115 N
c) 0.023 N
d) 0.046 N

40) The mole fraction of water in 20% (wt) aqueous solution of is
a) 77/68
b) 68/77
c) 20/80
d) 80/20

41) The molarity of 4% NaOH solution is
a) 0.1 M
b) 0.5 M
c) 0.01 M
d) 1.0 M

42) An aqueous solution of glucose is 10% in strength. The volume in which
1 gm mole of it is dissolved will be

a) 18 litres
b) 9 litres
c) 0.9 litres
d) 1.8 litres

43) If 5.85 g of NaCl are dissolved in 90 g of water the mole fraction of NaCl is
a) 0.1
b) 0.01
c) 0.2
d) 0.0196

44) How many grams of would have to be added to water to prepare 150 ml of a solution that is 2.0M ? (working)
a) 9.6
b) 2.4
c)
d)

45) The mole fraction of glycerine, in a solution containing 36 gm of water and 46 gm of glycerine is
a) 0.46
b) 0.40
c) 0.20
d) 0.36

46) A sugar syrup of weight 214.2 grams contains 34.2 grams of sugar (mol.wt. 342). The molal concentration is
a) 0.55
b) 5.5
c) 55
d) 0.1

47) A solution containing HCI molecules is diluted to a volume of 4 litres. The molar concentration of the solution is
a) 1
b) 2 M
c) 0.125 M
d) 0.25 M

48) In a solution of 7.8 g of benzene and 46 g of toluene , the mole fraction of benzene is
a) 1/6
b) 1/5
c) 1/2
d) 1/3

49) The mole fraction of glucose in an 18% solution of glucose is (working)
a) 0.18
b) 0.1
c) 0.017
d) 0.021

50) 120 g of urea is present in 5 L of solution. The active mass of urea is
a) 0.2
b) 0.06
c) 0.4
d) 0.88

51) The volume of 95% (density=1.85 g ) needed to prepare 100 of 15% solution of (density = 1.10 g ) will be
a) 5 cc
b) 7.5 cc
c) 9.4 cc
d) 12.4 cc

52) The molarity of the solution containing 7.1 g of in 100 ml of aqueous solution is
a) 2 M
b) 0.5 M
c) 1 M
d) 0.05 M

53) A solution of is 0.5 mol/litre, then the number of moles of chloride ions in 500 ml will be
a) 0.25
b) 0.50
c) 0.75
d) 1.00

54) The number of moles of hydroxide ion in 0.3 litre of 0.005 M solution of is
a) 0.0075
b) 0.0015
c) 0.0030
d) 0.0050

55) 150 ml of (density = 0.78 gm/ml) is diluted to one litre by adding water; molality of the solution is
a) 2.54
b) 11.7
c) 2.99
d) 29.9

56) What will be the molality of a solution having 18 g of glucose (molecular weight = 180) dissolved in 500 g of water?
a) 1 m
b) 0.5 m
c) 0.2 m
d) 2 m

57) 50 ml of 10 N, 25 ml of 12 N HCI and 40 ml of 5 N were mixed together and the volume of the mixture was made 1000 ml by adding water. The normality of the resulting solution will be
a) 1 N
b) 2 N
c) 3 N
d) 4 N

58) If moles of solute are dissolved in  moles of the solvent, Raoults law is expressed mathematically as
a)
b)
c)
d)

59) 5 ml of acetone is mixed with 100 ml of . The vapour pressure of water above the solution is
a) equal to the vapour pressure of pure water
b) equal to the vapour pressure of the solution
c) less than the vapour pressure of pure water
d) more than the vapour pressure of pure water

60) For determination of molecular weights, Raoults law is applicable only to
a) Dilute solutions of electrolytes
b) Concentrated solutions of electrolytes
c) Dilute solutions of non-electrolytes
d) Concentrated solution of non- electrolytes

61) The lowering of vapour pressure of the solvent takes place
a) only when the solute is non-volatile
b) only when the solute is volatile
c) only when the solute is a non-electrolyte
d) in all the above three cases

62) The lowering of vapour pressure of the solvent takes place on dissolving a non-volatile solute because
a) the density of the solution increases
b) the surface tension of the solution decreases
c) the viscosity of the solution increases
d) the molecules of the solvent on the surface are replaced by the molecules of the solute

63) A substance will be deliquescent if its vapour pressure is
a) equal to the atmospheric pressure
b) equal to that of the water vapour in the air
c) greater than that of the water vapour in the air
d) less than that of the water vapour in the air

64) Which of the following plots does not represent the behaviour of an ideal binary liquid solution?
a)
b)
c)
d)

65) For an ideal binary liquid solution with , which of the following relations between (mole fraction of A in liquid phase) and (mole fraction of A in vapour phase) is correctly represented?
a)
b)
c)
d)

66) The pressure under which liquid and vapour can coexist at equilibrium is called the
a) Limiting vapour pressure
b) Real vapour pressure
c) Normal vapour pressure
d) Saturated vapour pressure

67) The vapour pressure of a liquid in a closed container depends upon the
a) Amount of liquid
b) Surface area of the container
c) Temperature
d) None of the above

68) When attraction between A-B is more than that of A-A and B-B, the solution will show Â…Â…Â…. deviation from Raoults law:
a) positive
b) negative
c) no
d) unpredictable

69) Which of the following solution pairs can be separated into the pure components by fractional distillation?
a) Benzene – toluene
b)
c) Water Â– HCI
d)

70) Which of the following is correct for a solution showing positive deviations from Raoults law?
a)
b)
c)
d)

71) Pure water boils at 373 K and nitric acid boils at 359 K. The azeotropic mixture of and boils at 393.5 K. By distilling the azeotropic mixture
a) Pure nitric acid will distil over first
b) Pure water will distil over first
c) One of them, will distill over with a small amount of the other
d) Both of them will distil over in the same composition as that of the mixture being distilled.

72) The azeotropic mixture of water and HCI boils at . When this mixture is distilled, it is possible to obtain
a) pure hydrogen chloride
b) pure water
c) pure water as well as pure HCI
d)

73) The diagram given below is a vapour pressure composition diagram for a binary solution of A and B
In the solution, A-B interactions are

a) similar to A-A and B-B interactions
b) greater than A-A and B-B interactions
c) smaller than A-A and B-B interactions
d) unpredictable

74) Which pair from the following will not form an ideal solution?
a)
b)
c)
d)

75) When liquids A and B are mixed, hydrogen bonding occurs. The solution will show a
a)
b)
c)
d) Slight increase in volume

76) Which one of the following pairs will not form an ideal solution?
a)
b)
c)
d)

77) A binary solution of ethanol and n-heptane is an example of
a) an ideal solution
b) a non-ideal solution with positive deviation
c) a non-ideal solution with negative deviation
d) Unpredictable behaviour

78) For a dilute solution, Raoults law states that
a) The lowering of vapour pressure is equal to the mole fraction of the solute
b) the relative lowering of vapour pressure is equal to the mole fraction of the solute
c) The relative lowering of vapour pressure is equal to the amount of the solute in the solution
d) The vapour pressure of the solution is equal to the mole fraction of the solvent

79) Azeotropes are
a) liquid mixtures which distil unchanged in composition
b) Liquids which can mix with each other in all proportions
c) Solids which form solid solutions of definite composition
d) Gases which can be separated.

80) When a substance is dissolved in a solvent the vapour pressure of the solvent is decreased. This results in
a) an increase in the boiling point of the solution
b) a decrease in the boiling point of the solvent
c) the solution having a higher freezing point than the solvent
d) The solution having a lower osmotic pressure than the solvent

81) The boiling point of an azeotropic mixture of water and ethanol is less than that of water and ethanol. The mixture shows
a)
b)
c)
d) Deviations which cannot be predicted from the given information

82) A 0.6% solution of urea (molecular weight = 60) would be isotonic with
a) 0.1 M glucose
b) 0.1 M KCI
c) 0.6% glucose solution
d) 0.6% KCI solution

83) A mixture of benzene and toluene forms
a) an ideal solution
b) a non-ideal solution
c) a suspension
d) an emulsion

84) Which of the following liquid pairs shows a positive deviation from Raoults law?
a) Water-hydrochloric acid
b) Water-nitric acid
c) Acetone-chloroform
d) Benzene-methanol

85) Acetic acid dissolved in benzene shows a molecular mass of
a) 30
b) 60
c) 120
d) 180

86) The experimental molecular weight of an electrolyte will always be less than its calculated value because the value of the van’t Hoff factor i is
a) less than 1
b) greater than 1
c) equivalent to 1
d) zero

87) The law which states that the mass of a gas dissolved in a given mass of a solvent at any temperature is directly proportional to the pressure of the gas above the solution is called
a)
b)
c)
d)

88) According to Raoults law, the relative lowering of vapour pressure of solution is equal to the
a) Mole fraction of the solute
b) Number of moles of the solute
c) the mole fraction of the solvent
d) the number of moles of the solvent

89) Which one of the following is an expression of Raoults law if is the partial pressure of the solvent in a solution, is the partial pressure of the pure solvent and if and are the mole fraction of the solute and the solvent respectively?
a)
b)
c)
d)

90) If one mole of a substance is present in 1 kg of solvent, then its concentration is called the
a) Molar concentration
b) Molal concentration
c) Normality
d) Strength weight/weight

91) Which of the following is not correct regarding an ideal solution?
a)
b)
c)
d)

92) An azeotropic mixture of HCI and has
a) 48% HCI
b) 22.2% HCI
c) 36% HCI
d) 20.2% HCI

93) The vapour pressure of pure benzene and toluene are 160 and 60 torr respectively. The mole fraction of toluene in the vapour phase in contact with an equimolar solution of benzene and toluene is
a) 0.50
b) 0.6
c) 0.27
d) 0.73

94) The vapour pressure of a pure liquid A is 70 torr at . It forms an ideal solution with another liquid B. The mole fraction of B is 0.2 and the total vapour pressure of the solution is 84 torr at . The vapour pressure of pure liquid B at is
a) 14
b) 56
c) 140
d) 70

95) The vapour pressure of water at room temperature is 23.8 mm Hg. The vapour pressure of an aqueous solution of sucrose with the mole fraction 0.1 is equal to
a) 23.9 mm Hg
b) 24.2 Hg
c) 21.42 mm Hg
d) 21.44 mm Hg

96) The vapour pressure of is 143 mm Hg. 0.5 g of a non-volatile solute (mol. wt. 65) is dissolved in 100 ml of . Find the vapour pressure of the solution. (Density of =1.58
a) 141.93 mm
b) 94.39 mm
c) 199.34 mm
d) 143.99 mm

97) The vapour pressure of pure benzene and pure toluene at and 200 mm Hg respectively. If they form an ideal solution, what is the mole fraction of benzene in a mixture boiling at at a total pressure of 380 mm Hg?
a) 0.20
b) 0.40
c) 0.60
d) 0.80

98) The vapour pressure of pure A is 10.0 torr and at the same temperature when 1 g of B is dissolved in 20 g of A, its vapour pressure is reduced to 9.0 torr. It the molecular weight of A is 200 a.m.u., then the molecular weight of B is
a) 100 a.m.u.
b) 90 a.m.u.
c) 75 a.m.u.
d) 120 a.m.u.

99) 34.2 g of cane sugar is dissolved in 180 g of water. The relative lowering of vapour pressure will be
a) 0.0099
b) 1.1597
c) 0.840
d) 0.9901

100) The vapour pressure of benzene at a certain temperature is 640 mm Hg. A non-volatile non-electrolyte weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. What is the molecular weight of the solid substance?
a) 6.96
b) 65.3
c) 63.8
d) none of these

Ans 1) b

Ans Desc 1)

Ans 2) a

Ans Desc 2)

Ans 3) c

Ans Desc 3) ppm of F ions =0.2/500 x106 = 400

Ans 4) d

Ans Desc 4) 10% strength here means 10 g in 100 c.c. or 0.1 L. Hence volume containing 1 mol (i.e. 180 g) will be

Ans 5) a

Ans Desc 5)

Dividing
Taking

Ans 6) b

Ans Desc 6)

Ans 7) d

Ans Desc 7)

Ans 8) d

Ans Desc 8)

molecules

Ans 9) b

Ans Desc 9)

1m NaCl solution = 1 mol NaCl in 1000 g water
= 58.5 + 1000 g solution = 1058.5 g solution
= 1058.5/1.21 ml = 875 ml.

Ans 10) c

Ans Desc 10) 10% weight/volume means 10 g in 100 c.c.
Equivalent weight of

Ans 11) a

Ans Desc 11)

Ans 12) b

Ans Desc 12)

10 ml of 36 N = 10000 ml of ? N

Ans 13) d

Ans Desc 13) 100 ml of 0.3 N HCl contains HCl = 0.03 g eq.
200 ml of 0.6 N

Total g eq.=0.15, Total volume = 300 ml
Finally normality
Alternatively

Ans 14) b

Ans Desc 14)

Ans 15) b

Ans Desc 15)

Ans 16) a

Ans Desc 16) If the compound is impure, more than the theoretical weight Is required.

Ans 17) c

Ans Desc 17)

Hence molarity

Ans 18) b

Ans Desc 18)

Dividing
If

Ans 19) b

Ans Desc 19) 1000 ml of 1 N oxalic acid solution = 63 g
500 ml of 0.1 N oxalic acid solution = (63/1000) x 500 x 0.1
=3.15 g

Ans 20) a

Ans Desc 20)

Ans 21) b

Ans Desc 21) 1500 cc solution = 1500 x 1.052 g = 1578 g
Solvent = 1560 g, Solute = 18 g
Molality =

Ans 22) d

Ans Desc 22)

Ans 23) d

Ans Desc 23) Normality = Basicity x Molarity =3N

Ans 24) a

Ans Desc 24) Normality of oxalic acid sol.=
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Ans 25) a

Ans Desc 25) 98 g

Ans 26) a

Ans Desc 26)

3molal NaOH sol means 3 moles NaOH (120 g) in 1000 g water

Mass of solution = 1120 g
Volume of sol
Molarity

Ans 27) a

Ans Desc 27) 95% ethanol sol means 95 g ethanol in 100 g sol i.e. 5 g water

Ans 28) a

Ans Desc 28) Solvent = 188.8-10.6 = 178.2 g

Ans 29) b

Ans Desc 29) 0.2 mol i.e. 200 millimoles are present in 1000 c.c.

Ans 30) b

Ans Desc 30) 5.85 g NaCl = 0.1 mol.
As it is present in 1 kg of water, strength = 0.1molal.

Ans 31) c

Ans Desc 31)

Ans 32) c

Ans Desc 32)

Ans 33) c

Ans Desc 33) Molarity

Ans 34) c

Ans Desc 34)

Ans 35) b

Ans Desc 35) 60 ml of 4
=0.24 mol = 0.24×2 g eq.
=0.48 g eq. = 480 milli equivalents

Ans 36) a

Ans Desc 36) 5 molal aqueous ethyl alcohol solution means 5 moles of ethyl alcohol in 1000 g of water
= 1000/18 moles = 55.5 moles

Ans 37) a

Ans Desc 37) Molarity of pure water = 1000/18 = 55.6

Ans 38) d

Ans Desc 38) 18 g glucose = 18/180 mol=0.1 mol.
As it is present in 1000 g water, the solution is 0.1 molal.

Ans 39) a

Ans Desc 39) 0.115 g Na = 0.115/23 g eq. = 0.005 g eq.
Hence normality = 0.01

Ans 40) b

Ans Desc 40) 20% aq. solution means 20 g H2O2 in 100 g of solution
so that water = 80 g
hence the mole fraction of

Ans 41) d

Ans Desc 41) 4% NaOH means 4 g in 100 cc. Hence molarity = .

Ans 42) d

Ans Desc 42) 10% glucose sol means 10 g = 10/180 mole , or 1/18 mole in 100 cc i.e. 0.1 L
Hence 1 mole will be present in 18 x 0.1 = 1.8 L

Ans 43) d

Ans Desc 43) 5.85 g NaCl=

Ans 44) a

Ans Desc 44) 2M means 2 moles in 1000 cc solution
Hence the amount of in 150 ml
=0.3x32g=9.6g

Ans 45) c

Ans Desc 45)
46 g glycerine = 46/92=0.5 mole
\ mole fraction of glycerine = 0.5/2.5=0.20

Ans 46) a

Ans Desc 46) Water present = 214.2 – 34.2 = 180 g
Â Â Â Â Â Â 34.2 sugar = 0.1 mole
Â Â Â Â  Molal conc. = (0.1/180) x 1000 = 0.55

Ans 47) c

Ans Desc 47)
Molar conc. =

Ans 48) a

Ans Desc 48) 7.8 g benzene = 0.1 mole
46 g toluene = 0.5 mole
\ Mole fraction of benzene =

Ans 49) d

Ans Desc 49) Glucose = 18 g, solution = 100 g
\ Water = 82 g
Mole fraction of glucose =

Ans 50) c

Ans Desc 50) The active mass is the molar concentration
120 g urea
\ Active mass = 2/5 = 0.4 mole

Ans 51) c

Ans Desc 51) Molarity of 95%
Molarity of 15%

Ans 52) b

Ans Desc 52) Molarity

Ans 53) b

Ans Desc 53) 500 cc sol =0.25 mol

Ans 54) c

Ans Desc 54) 0.3 L of 0.005
=0.005 x 0.3 mol = 0.0015 mol

Ans 55) c

Ans Desc 55)
Water = 850 g
Molality =

Ans 56) c

Ans Desc 56) Molality =

Ans 57) a

Ans Desc 57)
50 x 10 + 25 x 12 + 40 x 5 =

Ans 58) a

Ans Desc 58) By Raoults law,

Ans 59) c

Ans Desc 59) Vapour pressure of any component in the solution = mole fraction of the component x vapour pressure of that component in the pure state and mole fraction is always <1.

Ans 60) c

Ans Desc 60) By Raoults law molecular weight is determined for dilute solutions of non-electrolytes.

Ans 61) d

Ans Desc 61) Lowering of vapour pressure of solvent takes place when the solute is either volatile or non-volatile,an electrolyte or a non-electrolyte.

Ans 62) d

Ans Desc 62)

Ans 63) d

Ans Desc 63) A substance is deliquescent if its V.P. is less than that of the water vapour in the air so that it can absorb water vapour from the air.

Ans 64) c

Ans Desc 64)

Ans 65) c

Ans Desc 65) As , A is more volatile and hence

Ans 66) d

Ans Desc 66) At equilibrium, the vapour pressure is the saturated vapour pressure.

Ans 67) c

Ans Desc 67) Vapour pressure of a liquid depends only on the temperature and the nature of the liquid.

Ans 68) b

Ans Desc 68) When A-B attractions are greater, less vapour is formed.

Ans 69) a

Ans Desc 69) Benzene + Toluene is an ideal solution. So its components can be separated by fractional distillation.

Ans 70) a

Ans Desc 70) For a non-ideal solution showing positive deviation,

Ans 71) d

Ans Desc 71) Azeotropic mixtures distil unchanged in composition.

Ans 72) d

Ans Desc 72) Azeotropic mixture cannot be separated into its components by simple distillation.

Ans 73) c

Ans Desc 73) As the solution shows positive deviation, A-B interactions are smaller.

Ans 74) b

Ans Desc 74) There is H-bonding between

Ans 75) b

Ans Desc 75) When H-bonding occurs on mixing, the solution shows negative deviation.

Ans 76) d

Ans Desc 76) Acetone is polar while is non-polar.

Ans 77) b

Ans Desc 77) There is H-bonding in ethyl alcohol, which is cut off on adding n-heptane. Hence it shows positive deviation.

Ans 78) b

Ans Desc 78) Raoults law states that the relative lowering of vapour pressure is equal to the mole fraction of the solute.

Ans 79) a

Ans Desc 79) Azeotropes distil unchanged in composition.

Ans 80) a

Ans Desc 80) The greater the lowering in vapour pressure, the higher is the boiling point.

Ans 81) b

Ans Desc 81) The boiling point of an azeotropic mixture is less when it shows positive deviations.

Ans 82) a

Ans Desc 82) 0.6% urea of glucose.

Ans 83) a

Ans Desc 83) Benzene and toluene are similar liquids

Ans 84) d

Ans Desc 84) Methanol has H-bonding. When benzene is added, H-bonds are cut off

Ans 85) c

Ans Desc 85) Acetic acid in benzene dimerizes to form

Ans 86) b

Ans Desc 86) An electrolyte dissociates. Hence i > 1

Ans 87) d

Ans Desc 87) Definition of Henrys law

Ans 88) a

Ans Desc 88) Raoults law states that the relative lowering of vapour pressure is equal to the mole fraction of the solute.

Ans 89) c

Ans Desc 89) By definition of Raoults law, . Note that is the mole fraction of the solvent in the solution)

Ans 90) b

Ans Desc 90) 1 mol of solute in 1 kg of solvent is a molal concentration

Ans 91) a

Ans Desc 91) For ideal solutions

Ans 92) d

Ans Desc 92) Azeotropic mixture of HCl and has 20.2% HCl.

Ans 93) c

Ans Desc 93) For equimolar solution,

Hence the mole fraction of toluene in the vapour phase = 30/110=0.27

Ans 94) c

Ans Desc 94)

Ans 95) c

Ans Desc 95)
or directly. Vapour pressure of solvent in the solution = mole of fraction of the solvent in the solution x vapour pressure of pure solvent.

Ans 96) a

Ans Desc 96)
or

Ans 97) c

Ans Desc 97)

Ans 98) b

Ans Desc 98)

or

Ans 99) a

Ans Desc 99)
Â Â Â Â Â Â Â Â Â Â Â Â Â

Ans 100) b

Ans Desc 100)
or
or
or M2=65.3

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