Test name: SACC-PS-CET-C-Solutions Standard: MHT-CET

Subjects: Chemistry Duration: 1Hr

Marks: 50

Questions

(C) 1.06 × 10–3 mol L–1

1. The molar mass of the solute using
depression of freezing point may be
calculated using the formula, (D) 2.60 × 10–3 mol L–1
_______.
4. In water saturated air, the mole
K f W 2 1000 fraction of water vapour is 0.02. If the
(A) M 2 =
total pressure of the saturated air is
ΔT f m
1.2 atm, the partial pressure of dry
(B) M 2 =
K f W 1 1000 air is _______. [NEET (Odisha)
ΔT f W 2 2019]

(C) M 2 =
ΔT f W 2 1000
(A) 0.98 atm
Kf W1

(B) 1.18 atm

K f W 2 1000
(D) M 2 =
ΔT f W 1
(C) 1.76 atm
2. Which has the minimum freezing
point? (D) 1.176 atm

(A) One molal NaCl solution 5. The colligative properties of a

solution depend on _______.
(B) One molal KCl solution
(A) nature of solute particles present
(C) One molal CaCl2 solution in it

(B) nature of solvent used

(D) One molal urea solution
(C) number of solute particles
3. The solubility of nitrogen gas in
present in it
water at 2 bar pressure at 25 °C is
13.6 × 10–4 mol L–1. Calculate the (D) number of moles of solvent only
solubility of N2 gas in water at 25 °C,
when the partial pressure of N2 gas 6. The value of Kb for a solvent is X K
in the atmosphere is 1.56 bar. kg mol–1. A 0.2 m solution of a non-
electrolyte in this solvent will boil at
(A) 1.60 × 10–3 mol L–1 _______. (Given: boiling point of
solvent = A °C).
(B) 2.16 × 10–3 mol L–1 (A) (A + X) °C
 (B) (A+ ) °C Kf for benzene is 5.12 K kg mol–1.
X
10

(C) (A + ) °C (A) 256 g mol–1

X
5

(D) (A + X
)K (B) 2.56 g mol–1
0.2

7. At constant temperature, the osmotic (C) 512 g mol–1

pressure of a solution is _______.
(D) 2.56 × 104 g mol–1
(A) directly proportional to the
concentration 11. Which of the following statements is
INCORRECT?
(B) inversely proportional to the
concentration (A) NaCl dissolves in water.
(C) directly proportional to the (B) Cholesterol dissolves in
square of the concentration benzene.
(D) directly proportional to the (C) Sugar does not dissolve in water.
square root of the concentration
(D) Naphthalene dissolves in
8. Aqueous solutions containing 1.63 g benzene.
of boric acid in 450 mL and 20 g of
sucrose (molecular mass = 342) per 12. If the elevation in boiling point of a
litre are isotonic. The molar mass of solution of 50 g of solute (molar
boric acid is _______. mass = 100) in 500 g of water is
ΔTb. The ebullioscopic constant Kb
(A) 342×1.63
20 of water is equal to _______. [MHT
CET 2019]
1.63×1000×342
(B) 20×450
(A) 100 ΔTb
(C) 1.63×342×450
ΔT b
1000×20
(B)
50

(D) 20×342×450
1000×1.63 (C) 10 ΔTb
9. 9 g of glucose (mol wt = 180) is (D) ΔTb
dissolved in 90 g of H2O. Relative
lowering of vapour pressure is 13. Which is CORRECT about Henry’s
_______. [MH CET 2011] law?
(A) 0.99 (A) Solubility of a gas in a liquid
decreases with decrease in
(B) 0.099
temperature.
(C) 0.0099 (B) Solubility of a gas in a liquid
(D) 0.00099 increases with increase in external
pressure.
10. 1.00 g of a non-electrolyte solute
dissolved in 50 g of benzene lowered (C) Solubility of a gas in a liquid
the freezing point of benzene by 0.40 decreases with increase in external
pressure.
K. Molecular mass of the solute will
be _______.
 (D) Solubility of a gas in a liquid 18. Vapour pressure of the solution, of a
increases with increase in non-volatile solute is always
temperature and decrease in _______.
external pressure.
(A) equal to the vapour pressure of
14. Which of the following shows pure solvent
negative deviation from Raoult’s
law? (B) higher than vapour pressure of
pure solvent
(A) Benzene + toluene
(C) lower than vapour pressure of
(B) Phenol + aniline pure solvent

(C) Ethanol + acetone (D) constant

(D) Carbon disulphide + acetone 19. Depending on the physical states of

solvents and solutes, there are
15. The solubility of a gas in water _______ types of solutions.
increases with _______.
(A) five
(A) increase in temperature
(B) seven
(B) reduction in gas pressure
(C) nine
(C) decrease in temperature
(D) twelve
(D) amount of liquid taken
20. 18 g glucose (C6H12O6) is added to
16. At 300 K, when a solute is added to 178.2 g water. The vapour pressure
a solvent, its vapour pressure over of water (in torr) for this aqueous
mercury reduces from 50 mm to 45 solution is _______. [JEE (Main)
mm. The value of mole fraction of 2016]
solute will be _______.
(A) 7.6
(A) 0.005
(B) 76.0
(B) 0.01
(C) 752.4
(C) 0.1
(D) 759.0
(D) 0.9
21. 18 g of glucose is dissolved in 90 g
17. The depression of freezing point of of H2O. Relative lowering of vapour
water for a particular solution is
0.186 K. The boiling point of the pressure is _______. [Molar mass of
same solution is _______. glucose = 180]

(A) 0.0196
(Kf = 1.86 K kg mol–1, Kb = 0.512 K
(B) 0.180
kg mol–1)
(C) 0.0990
(A) 100.186 K
(D) 0.510
(B) 100.512 K
22. Calculate osmotic pressure of
(C) 273.512 K
solution of 0.025 mole glucose in
(D) 373.0512 K 100 mL water at 300 K.
 [R = 0.082 atm dm3 mol–1 K–1] (D) the sum of vapour pressure of
pure solvent and mole fraction of the
MHTCET-2023 pure solvent

(A) 1.54 atm 26. During osmosis, flow of water

through a semipermeable membrane
(B) 2.05 atm is _______.

(C) 6.15 atm (A) unidirectional

(D) 3.08 atm (B) bidirectional

23. If molality of the dilute solution is (C) multidirectional

doubled, the value of molal
depression constant (Kf) will be (D) unpredictable
_______. [NEET (UG) 2017] 27. The amount of urea to be dissolved
in 500 g of water to produce a
(A) halved
depression of 0.186 K in the freezing
(B) tripled point is _______.

(C) unchanged (Kf for water = 1.86 K kg mol–1)

(D) doubled
(Molecular mass = 60 g mol–1)
24. How much polystyrene of molecular (A) 0.6 g
mass 9000 g mol–1 would have to
be dissolved in 100 g of C6H6 to (B) 60 g
lower its freezing point by 1.05 K? (C) 3 g

(Kf of C6H6 = 4.9 K kg mol–1). (D) 6 g

(A) 19.3 g 28. van’t Hoff factor is _______.

(B) 193 g (A) less than one in case of

dissociation
(C) 38.6 g
(B) always more than one
(D) 77.2 g
(C) always less than one
25. The lowering of vapour pressure of a
solvent by the addition of a non- (D) greater than one in case of
volatile solute to it, is equal to dissociation
_______.
29. 6 g of substance x dissolved in 100 g
(A) the product of vapour pressure of of water lowers the freezing point by
pure solvent and mole fraction of the 0.93 K. The molar mass of x is
non-volatile solute _______.

(B) the product of vapour pressure of (Kf = 1.86 K kg mol–1)

pure solvent and mole fraction of the
pure solvent
(A) 60 g mol–1
(C) the sum of vapour pressure of
pure solvent and mole fraction of the (B) 120 g mol–1
non-volatile solute
 (C) 180 g mol–1 34. Vapour pressure of pure solvent and
its solution at certain temperature
are 660 mm and 600 mm of Hg
(D) 140 g mol–1
respectively. If 3.6 × 10–3 kg of
30. The mass of ascorbic acid (C6H8O6) solute is added into 40 × 10–3 kg of
to be dissolved in 100 g of acetic solvent, what is the molar mass of
acid to lower its freezing point by 1.5 solute? (solvent = Benzene, C = 12,
°C in g is _______. H = 1) [MH CET 2013]

(Kf of acetic acid = 4 K kg mol–1) (A) 78.0 g mol–1

(A) 0.015 (B) 58.5 g mol–1

(B) 6.6
(C) 72.0 g mol–1
(C) 0.30
(D) 156 g mol–1
(D) 0.66
35. Which of the following solution has
31. At 100 °C, the vapour pressure of a highest osmotic pressure?
solution of 6.5 g of a solute in 100 g
water is 732 mm. If Kb = 0.52, the (A) 1 M NaCl
boiling point of this solution will be (B) 1 M urea
_______. [NEET P-I 2016]
(C) 1 M sucrose
(A) 102 °C
(D) 1 M glucose
(B) 103 °C
36. Benzene and toluene form nearly
(C) 101 °C ideal solutions. At 20 °C, the vapour
(D) 100 °C pressure of benzene is 75 torr and
that of toluene is 22 torr. The partial
32. Which of the following is a colligative vapour pressure of benzene at 20 °C
property? for a solution containing 78 g of
benzene and 46 g of toluene in torr
(A) Melting point of a substance is _______.

(B) Surface tension of a solution (A) 50

(C) Boiling point elevation (B) 25

(D) Radioactivity of a substance (C) 37.5

33. 0.1 m solution each of glucose, (D) 53.5

sodium chloride, sodium sulphate
and sodium phosphate are taken; 37. Two elements X and Y form
the ratio of depression in freezing compounds having molecular
point is _______. formula XY2 and XY4. When
dissolved in 20 g of benzene, 1 g of
(A) 1 : 3 : 2 : 1 XY2 lowers the freezing point by 2.3
K, whereas 1.0 g of XY4 lowers it by
(B) 1 : 2 : 3 : 4
1.3 K. The molal depression
(C) 4 : 3 : 2 : 1 constant for benzene is 5.1 K kg

(D) 1 : 1 : 2 : 4
 mol–1. The atomic mass of X and Y (A) 22800×R
1.5
is _______ respectively.
22800
(B)
(A) 23.88, 40.31 1.5

(B) 25.59, 42.64 (C) 1.5×R

22800

(C) 27.13, 44.83

(D) 228×R
1.5

(D) 29.28, 46.73

42. The boiling point of 0.15 molal
38. 1 M and 2 M solutions of glucose are aqueous solution of an unknown
prepared in water. Hence, _______. solute is 373.23 K at 1 atm. The
molal elevation constant of water is
(A) the osmotic pressure of both the _______ K kg mol–1.
solutions will be the same at the
same temperature (A) 0.53
(B) 2 M solution will have higher (B) 0.88
osmotic pressure
(C) 1.8
(C) 1 M solution will have higher
osmotic pressure (D) 5.3
(D) osmotic pressure will be 43. A solution containing 3.56 g of a
independent of the concentration polymer in 1 litre of a solvent was
found to have an osmotic pressure of
39. True solutions contain solute
particles with diameters in the range 5.2 × 10–4 atmosphere at 300 K.
of _______ nm. The molecular mass of the polymer
is _______.
(A) 0.1 to 2
(R = 0.082 L atm mol–1 K–1)
(B) 1 to 20
(A) 1.68 × 103 g mol–1
(C) 10 to 200

(D) 100 to 2000 (B) 1.68 × 102 g mol–1

40. Which of the following solutions (C) 1.68 g mol–1

exhibits positive deviation from
Raoult’s law? (D) 1.68 × 105 g mol–1
(A) Chloroform + Acetone 44. The vapour pressure of benzene at
80 °C is lowered by 10 mm by
(B) Ethanol + Acetone
dissolving 2 g of a non-volatile
(C) Aniline + Phenol substance in 78 g of benzene. The
vapour pressure of pure benzene at
(D) Benzene + Toluene 80 °C is 750 mm. The molecular
mass of the substance will be
41. If 10 g of a solute was dissolved in _______ g mol–1.
250 mL of water and osmotic
pressure of the solution was found to (A) 15
be 600 mm of Hg at 300 K, then
molecular weight of the solute is (B) 150
_______ g mol–1.
(C) 1500
 (D) 2000 water at a particular temperature is
2985 N/m2. The vapour pressure of
45. The ratio of the value of any
colligative property for KCl solution pure water is 3000 N/m2. The molar
to that of a sugar solution is nearly mass of the solute is _______.
_______.
(A) 60 g mol–1
(A) 1
(B) 120 g mol–1
(B) 0.5

(C) 2.0 (C) 180 g mol–1

(D) 3 (D) 380 g mol–1

46. An example for solid in solid solution 49. If two solutions separated by a
is _______. semipermeable membrane have the
same osmotic pressure, they are
(A) bronze called _______ solutions.
(B) mercury amalgam (A) hypertonic
(C) sodium chloride in water (B) hypotonic
(D) camphor in air (C) isotonic
47. Select the CORRECT option for a (D) saturated
solution containing non-volatile
solute in it. 50. The molal freezing point depression
0 constant for water is 1.86 K kg mol–
(A) T f < Tf 1. If 342 g of cane sugar
0
(C12H22O11) is dissolved in 1000 g
(B) T b < Tb of water, the solution will freeze at
_______.
0
(C) P = P 1 x1
(A) – 1.86 °C
0
(D) Tf = Tf – T f (B) 1.86 °C

48. The vapour pressure of a solution of (C) – 3.92 °C

5 g of non-electrolyte in 100 g of
(D) 2.42 °C