Chemistry Questions and Answers – Colligative Properties and Determination of Molar Mass – 2


This set of Chemistry Online Quiz for Schools focuses on “Colligative Properties and Determination of Molar Mass – 2”.

1. Why is ‘raising of viscosity’ of a solution after addition of solute, not considered to be a colligative property?
a) The resultant viscosity depends on the nature of the solute
b) The resultant viscosity depends on the amount of solute
c) The resultant viscosity depends on the nature of solvent
d) The resultant viscosity depends on the amount of solvent
View Answer

Answer: a
Explanation: A colligative property is identified by the fact that it has no dependence on the nature of the particles of solute. However, in the case of viscosity it really depends on the solute that is added to the solvent. Thus, the change in viscosity after addition of a non-volatile solute cannot be considered to be a colligative property.

2. At 70°C the vapor pressure of pure water is 31 kPa. Which of the following is most likely the vapor pressure of a 2.0 molal aq. glucose solution at 70°C?
a) 30.001 kPa
b) 29.915 kPa
c) 28.226 kPa
d) 32.392 kPa
View Answer

Answer: b
Explanation: Given, P0water = 31 kPa
Concentration of solution, c = 2 molal = 2 moles of glucose/kg of water
From law of relative lowering of vapor pressure, ΔP/P0 = X2, where X2 is the mole fraction of glucose in the solution.
Mass of water = 1 kg = 1000 g
Molecular weight of water = 18 g/mole
Moles of water = 1000/18 = 55.556 moles
X2 = 2/(2 + 55.556) = 0.035
ΔP = 31 kPa x 0.035 = 1.085 kPa
Final pressure = 31 kPa – 1.085 kPa = 29.915 kPa.

3. 117 g of NaCl is added to 222 g of water in a saucepan. At what does temperature does water boil at 101.325 kPa? Ebullioscopy constant for water = 0.52 K kg mol-1 and b.p. = 100°C
a) 98.3°C
b) 102.8°C
c) 104.7°C
d) 101.5°C
View Answer

Answer: c
Explanation: Given,
Weight of solvent, w1 = 222 g
Weight of solute, w2 = 117 g
Kb = 0.53 K kg mol-1
Now, addition of a non-volatile solute causes elevation in boiling point, ΔTb
ΔTb = (kb x 1000 x w2)/(M2 x w1)
On substituting, ΔTb = (0.52 x 1000 x 117)/(58.5 x 222) = 4.7°C
New boiling temperature = 100 + 4.7 = 104.7°C.

4. Boiling point of chloroform is 61°C. After addition of 5.0 g of a non-volatile solute to 20 g chloroform boils at 64.63°C. If kb = 3.63 K kg mol-1, what is the molecular weight of the solute?
a) 320 g/mol
b) 100 g/mol
c) 400 g/mol
d) 250 g/mol
View Answer

Answer: d
Explanation: Given,
b.p. of chloroform = 61°C
New b.p. after addition = 64.63°C
Mass of solute, w2 = 5.0 g
Mass of solvent, w1 = 20 g
Kb = 3.63 K kg mol-1
From these, ∆Tb = 64.63 – 61 = 3.63°C
Using ΔTb = (kb x 1000 x w2)/(M2 x w1)
M2 = (kb x 1000 x w2)/(ΔTb x w1)
M2 = (3.63 x 1000 x 5)/(3.63 x 20) = 250 g/mol.

5. Pure CS2 melts at -112°C. 228 grams of propylene glycol crystals is mixed with 500 grams of CS2. If kf of CS2 = -3.83 K kg mol-1 what is the depression in freezing point?
a) -23°C
b) -135°C
c) -20°C
d) -100°C
View Answer

Answer: a
Explanation: Given,
kf = -3.83 k kg mol-1
Mass of solute, w2 = 228 g
Mass of solvent, w1 = 500 g
Molar mass of solute, M2 = 76 g/mole
Moles of solute = w2/M2 = 228/76 = 3 moles
Molality of the solution, m = Number of moles of solute/mass of solvent (kg)
m = 3 moles/0.5 kg = 6 molal
We know, ΔTf = kf x m
ΔTf = -3.83 x 6 = -23°C.

6. What are colligative properties useful for?
a) Determining boiling and melting temperature
b) Determining molar mass
c) Determining equivalent weight
d) Determining van’t Hoff factor
View Answer

Answer: b
Explanation: Colligative properties serve the purpose for determining molar masses of unknown compounds. Colligative properties are not used to determine boiling and melting temperatures as it would result in an incorrect value upon the addition of a solute. Equivalent weight can only be determined if the molar mass is known and the van’t Hoff factor is determined in a similar manner.

7. A pair of solution bears the same osmotic pressure. What is this pair of solutions called?
a) Hypertonic
b) Hypotonic
c) Isotonic
d) Osmolarity
View Answer

Answer: c
Explanation: Two solutions having the same osmotic pressure at a given temperature are called isotonic solutions. Hypertonic solutions are those in which the concentration of solute outside the membrane and a lesser concentration within the membrane. Osmolarity is a type of concentration expressed in number of solute particles per liter.

8. A cell with lots of salt inside it is placed in a vessel containing just water. Which process takes place?
a) Dialysis
b) Filtration
c) Shriveling
d) Osmosis
View Answer

Answer: d
Explanation: Osmosis is a mass transfer process due to which water molecules move from a region of higher water potential to lower water potential down the potential gradient. Dialysis and filtration are processes using the concept of diffusion of impurities. Shriveling is the shrinking of a cell. In this case, the cell swells up.

9. What is a necessary condition for osmosis to take place?
a) Semi-permeable membrane
b) Same concentration of solvent
c) High temperature
d) Pressure greater than osmotic pressure
View Answer

Answer: a
Explanation: A semi permeable membrane is a must condition since it facilitates the blocking of solute particles from diffusing through and allows only water molecules to pass through. Obviously, water is the solvent in osmosis hence the concentration of solvent cannot be same if osmosis has to occur. If pressure greater than osmotic pressure is applied, reverse osmosis takes place. Meaning, water molecules will flow from region of lower water potential to higher water potential.

10. Which is the most appropriate method for determining the molar masses of biomolecules?
a) Relative lowering of vapor pressure
b) Elevation of boiling point
c) Depression in freezing point
d) Osmosis
View Answer

Answer: d
Explanation: Osmotic pressure method has the greatest advantage over other methods that even for very dilute concentrations it gives a large magnitude. This would not be true in case other methods. It is highly useful for biomolecules since they are unstable at extremely high and low temperatures thus eradicating the method using boiling and freezing points.

Sanfoundry Global Education & Learning Series – Chemistry – Class 12.

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