Question

What weight of glycerol should be added to 600g of water in order to lower its freezing point by ${10^0}C$?
Given that: ${K_f} = {1.86^0}C{m^{ - 1}}$
(A) 496g
(B) 297g
(C) 310g
(D) 426g

Answer 1

Hint: In order to solve this question, you must be aware about the concept of freezing point depression. The freezing point depression of a solution is a colligative property of the solution which is dependent upon the number of dissolved particles in the solution. The higher the solute concentration, the greater the freezing point depression of the solution.

Complete answer: Given that:
$\Delta {T_f} = {10^0}C$ ($\Delta {T_f}$ represents the freezing point of pure solvent)
Now,
$\Delta {T_f} = {K_f} \times molality $(${K_f}$ represents Freezing point depression constant)
Molality is defined as the moles of solute per kg of the solvent.
$10 = 1.86 \times \dfrac{{x \times 1000}}{{92 \times 600}}$ (Molecular mass of glycerol is 92g)
$x = \dfrac{{10 \times 600 \times 92}}{{1.86 \times 1000}}$
$x = 296.8g$
Wt. of glycerol taken $ \approx $297g
Therefore, Option (B) is correct.

Note:
The freezing point of the pure solvent is at constant temperature but the freezing point of the solution slowly decreases. The decrease is caused by the increase in solute concentration as the solvent freezes. The dissolved solutes can be non-electrolytes or electrolytes. Non-electrolytes are molecules that remain intact when they dissolve in water. Electrolytes are solutes that dissociate into ions when dissolved in solution to give an electrically conducting solution.