Chemistry Online Quiz for Schools 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
Answer: a
Clarification: 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
Answer: b
Clarification: 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
Answer: c
Clarification: 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
Answer: d
Clarification: 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
Answer: a
Clarification: 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