Question

Let’s think that you are driving in a car in winter. In a cold climate water gets frozen causing damage to radiator of a car. Ethylene glycol is used as an antifreezing agent. Calculate the amount of ethylene glycol to be added to 4 kg of water to prevent it from freezing at $–6^{\circ }C$. $(K_f$ for water = $1.85Kkgmo{l}^{-1})$

• A. 804.32g
• B. 904.32 g
• C. 704.32 g
• D. 1004.32 g

Answer 1

The correct option is A 804.32g

So bad that you are stuck with damaged radiator of the car.
Chemistry comes to rescue here.
As given in the question ethylene glycol can be used as a anti freezing agent.
We can calculate the amount of ethylene glycol to be added by our knowledge of depression in freezing point. Let’s check the given information.
$\Delta T=6^{\circ }C,W_1=4kg=4000g,M{w}_2=62,K_f$ (molal depression constant) = 1.85
We just saw how to calculate mass of ethylene glycol here.
$W_2=\frac{M{w}_2\times W_1\times \Delta T}{1000\times K_f}$
$\Delta T$ = Depression in freezing point
$K_f$ = Molal depression constant of benzene
$W_2$ = Mass of solute
$M{w}_2$ = Molecular mass of solute
$W_1$ = Mass of solvent
$\frac{62\times 4000\times 6}{1000\times 1.85}=804.32g$