Question

$K_f$ for water is $1.86$ $kgmo{l}^{-1}$.If your automobile radiator holds in 1.0kg of water, how many grams of ethylene glycol $(C_2{H}_6{O}_2)$ must you add to get the freezing point of the solution lowered to $-2.8{}^{\circ }$C.
(A) 72 g
(B) 93 g
(C) 39 g
(D) 27 g

Answer 1

Hint: freezing point of water is $0{}^{\circ }$C. depression of freezing point is 2.8 degree Celsius. Formula to calculate grams of ethylene glycol is as follows: $\mathrm{\Delta }{T}_f=K_f{\displaystyle \frac{w_2\times 1000}{M_2{w}_1}}$
The molar mass of ethylene glycol is 62.

Complete step by step solution:
-Vapour pressure of pure water will be lowered when a solute is added. As more solute is added, vapour pressure goes on lowering. Raoult’s Law states that partial vapour pressure of each component is directly proportional to its mole fraction present in the component.
-When a non-volatile solute is added to the solvent, there is a decrease in vapour pressure and it becomes equal to that of solid solvent at a lower temperature so there is a decrease in freezing point of the solvent.
-Freezing point of water is 0 degree Celsius, required freezing point after addition of ethylene glycol is -2.8 degree Celsius, so depression is the freezing point is 2.8 degree Celsius.
-Depression of freezing point is directly proportional to molality.
$\mathrm{\Delta }{T}_f\propto m$
So $$\mathrm{\Delta }{T}_f=K_{f}m$$
$K_f$ is cryoscopic constant or molal depression constant. It is defined as depression in freezing point for a molal solution.
-$$\mathrm{\Delta }{T}_f=K_f{\displaystyle \frac{w_2\times 1000}{M_2{w}_1}}$$
$2.8{}^{\circ }\text{C= }{\displaystyle \frac{\text{1}\text{.86}\times {\text{w}}_2\times 1000}{62\times 1000}}$
$w_2$ =93g

Option (B) 93 grams of ethylene glycol $(C_2{H}_6{O}_2)$ must you add to get the freezing point of the solution lowered to $-2.8{}^{\circ }$C.

Note: molal depression constant is a depression in freezing point when one mole of solute is added to 1kg of solvent. Depression in freezing point can be calculated by taking the difference between the freezing point of pure solvent and freezing point of the solution after the addition of solute. depression of freezing point is directly proportional to molality.