Chemistry Questions and Answers – Colligative Properties and Determination of Molar Mass – 1

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This set of Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Colligative Properties and Determination of Molar Mass – 1”.

1. What are the properties arising due to varying concentrations of solute in a given solvent, irrespective of the nature of solute with respect to the solvent?
a) Colligative properties
b) Intensive properties
c) Extensive properties
d) Solute properties
View Answer

Answer: a
Explanation: Colligative properties are set of four properties. This set of properties purely depends on the number of solute particles present in the solution/solvent, independent of the nature of the particles with respect to the solution/solvent. The properties are 1.) Relative lowering of vapor pressure, 2.) Elevation in boiling point, 3.) Depression in freezing point and 4.) Osmotic pressure.
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2. Which of the following is a colligative property?
a) Relative lowering of fluid pressure
b) Decrease in boiling point
c) Decrease in freezing point
d) Change in volume after mixing
View Answer

Answer: c
Explanation: Decrease in freezing point is the correct colligative property, known as ‘Depression in freezing point’. This is caused by solute particles present on the surface which lowers the equilibrium solid-vapor pressure. Therefore, a lower freezing temperature is required to match the pressure outside. The other correct colligative properties are 1.) Relative lowering of vapor pressure, 2.) Elevation in boiling point, 3.) Depression in freezing point and 4.) Osmotic pressure.

3. Which law specifically governs the relative lowering of vapor pressures in solutions?
a) van’t Hoff law
b) Boyle’s law
c) Raoult’s law
d) Amagat’s law
View Answer

Answer: c
Explanation: Raoult’s law quantifies the relative lowering in vapor pressure. From the law, it follows that p1 = X1 x p01. If p01 was the original pressure before X2 mole fraction of solute was added to the solvent then reduction in vapor pressure is given as:
Δp1 = p01 – p1
= p01 – p01 X1
= p01(1 – X1)
= p01X2
Therefore, the relative reduction in vapor pressure (∆p1/p01) = X2 i.e. mole fraction of solute in the solution.
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4. If 1.5 grams of a non-volatile solute (MW = 100) is added to 200 ml pure CS2 (ρ = 1.3 g/cc) whose vapor pressure is 400 mm of Hg at 28.0°C, what is the resulting vapor pressure of the dilute solution?
a) 401.246 mm Hg
b) 398.754 mm Hg
c) 401.754 mm Hg
d) 398.246 mm Hg
View Answer

Answer: d
Explanation: Given,
Mass of solute, m2 = 1.5 grams
Molar mass of solute, M2 = 100
Vapor pressure of pure CS2, p01 = 400 mm Hg
Volume of solvent, ρ = 200 ml
Density of solvent, ρ = 1.3 g/cc
Number of moles of solute, n2 = m2/MW = 1.5/100 = 0.015 mole
From the law of relative lowering of vapor pressure, Δp1 = p01X2, where X2 is the mole fraction of solute and Δp1 is the difference in pressure.
Mass of solvent, m1 = ρ x V = 1.3 x 200 = 260 grams
Number of moles of solvent, n2 = 260g/[(12 + 32 + 32)g/mole] = 3.421
Since the solution is dilute we can approximate X2 = n2/(n1 + n2) ≈ n2/n1 (since n2<< n1)
Using ∆p1 = p01X2,
∆p1 = (400 mg of Hg) (0.015/3.421) = 1.754 mm of Hg
Using Δp1 = p01 – p1, we get p1 = p01 – Δp1
Hence, resulting lowered vapor pressure, p1 = 400 mm Hg – 1.754 mm Hg = 398.246 mm Hg.

5. If the relative lowering of pressure of o-xylene is 0.005 due to addition of 0.5 grams of non-volatile solute in 500 grams of solvent, what is the molecular weight of the solute?
a) 21.3 g/mole
b) 23.1 g/mole
c) 32.1 g/mole
d) 1.23 g/mole
View Answer

Answer: a
Explanation: Given,
∆p1/p01 = 0.005
Mass of solute, mW = 0.5 grams
Mass of solvent, mS = 500 grams
Number of moles of solute, nS = 0.5/MW, where MW is the molecular weight of the solute.
Number of moles of solvent, nSolvent = 500 grams/(106 g/mole) = 4.7 mole
Since mS<<mSolvent , the solution can be considered to be dilute. Hence, equation for relative lowering for vapor pressure becomes:
Δp1/p01 = n2/(n1 + n2) ≈ n2/n1
On substituting values,
0.005 = (0.5/MW)/4.7
On solving, MW = 21.3 g/mole.
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6. On addition of non-volatile potassium iodide in water at 298K it is noticed that vapor pressure reduces from 23.8 mm Hg to 2.0 cm Hg. What is the mole fraction of solute in the solution?
a) 0.916
b) 0.160
c) 0.084
d) 0.092
View Answer

Answer: b
Explanation: Given,
P0water =23.8 mm Hg
Pwater= 2.0 cm Hg = 20.0 mm Hg (after addition of solute)
From the law of relative lowering of vapor pressure, ∆p1 = X2 x p01 (where X2 is the mole fraction of solute)
On rearranging, Δp1/p01 =X2
Δp1 = 23.8 – 20.0 = 3.8 mm Hg
X2 = 3.8/23.8 = 0.160.

7. When a non-volatile solute is added to a solvent what is the difference in vapor pressure expressed as a faction of original vapor pressure equal to?
a) Mole fraction of solute in vapor phase
b) Mole fraction of solvent in vapor phase
c) Mole fraction of solute in liquid phase
d) Mole fraction of solvent in liquid phase
View Answer

Answer: c
Explanation: Relative lowering of vapor pressure is the difference in vapor pressure expressed as a fraction of original vapor pressure. Using Raoult’s law:
ΔP1 = X2 x P1
Therefore, ΔP1/P1 = X2 which implies the relative lowering is equal to mole fraction of solute in liquid phase. Since the solute is non-volatile, it cannot be present in vapor phase.
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8. When 2.0 grams of copper (II) nitrate is added to 1000 ml of pure water, by how much is the vapor pressure of water decreased, given that at 20°C the vapor pressure of pure water is 17.535 mm Hg?
a) 3.1 x 10-4
b) 0.303
c) 0.0333
d) 0.0033
View Answer

Answer: d
Explanation: Given,
Mass of solute, msolute = 2.0 grams
Volume of solvent, Vsolvent = 1000 ml
P0water = 17.535 mm Hg
Molecular mass of Cu(NO3)2 = 188 g/mole
We know that ΔP1/P01 = X2, where X2 is the mole fraction of the solute.
We are required to find out the difference in vapor pressure i.e. ΔP1.
Therefore, ΔPwater= P0water x X2
Number of moles of copper (II) nitrate = 2.0/188 = 0.0106
Number of moles of water = 1000g/(18g/mole) = 55.56 mole (since density of water is 1g/ml)
Mole fraction of solute, X2 = 0.0106/(0.0106 + 55.56) = 1.9 x 10-4
ΔPwater = 17.535 x 1.9 x 10-4 = 0.0033.

9. Two components A and B have their pure vapor pressures in the ratio 1 ∶ 4 and respective mole fractions in solution in ratio 1 ∶ 2. What is the mole fraction of component B in vapor phase?
a) 0.8889
b) 0.1250
c) 0.8000
d) 0.2000
View Answer

Answer: a
Explanation: Given,
4P0A = P0A ———- (1)
2XA = XB ———— (2)
Using law of relative lowering of vapor pressure:
PA = P0A x XA ——- (3)
PB = P0B x XB ——- (4)
Using values of (1) and (2) in (4)
PA = P0A x XA
PB = 4P0Ax 2XA =8(P0A x XA)
Therefore, PB = 8PA
Mole fraction of component B in vapor phase, YB = PB/(PA + PB)
On simplifying, YB = 8PA/(PA + 8PA) = 8/9 = 0.8889.
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10. Which of the following is Raoult’s law applicable to, in order to determine molar masses correctly?
a) Ionic solute in liquid
b) Non-ionic solute in dilute solution
c) Non-ionic solute in concentrated solution
d) Ionic solid in insoluble form in solvent
View Answer

Answer: b
Explanation: In order to determine molar masses correctly, it is essential that the solute is non-volatile, non-ionic and present in dilute form only. If it is ionic one has to account for its van’t Hoff factor, i, for association in concentrated solution and dissociation in dilute solution. Molar masses can only be calculated from dilute solutions containing non-dissociable non-ionic solutes.

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