Question

In cold climates, water gets frozen causing damage to the radiator of a car. Ethylene glycol is used as an anti-freezing agent. Calculate the amount of ethylene glycol to be added to $4Kg$ of water to prevent it from freezing at $- 6^\circ C$ given that ${K_f}$ of water is $1.85KKgMo{l^{ - 1}}$.

Answer 1

When a solute is added to pure water, its vapor pressure will be lowered. This lowered vapor pressure is equal to the vapor pressure of the solute added and hence the freezing point of water decreases.

Complete step-by-step answer:
Depression of freezing point is a colligative property that occurs in solutions as a result of addition of solute molecules to a solvent. The depression in freezing point is directly proportional to the molarity of the solute.
The normal freezing point of water is $0^\circ C$
Given to us, the freezing point to is $- 6^\circ C$
Hence, the depression of freezing point is $\Delta {T_f} = 0 - \left( { - 6} \right) = 0 + 6 = 6^\circ C$
Now, let us calculate the molarity of ethylene glycol.
Molar weight of ethylene glycol is $62$
Let the required mass be x.
Then the molarity would be $M = \dfrac{x}{{62}} \times \dfrac{1}{4} = \dfrac{x}{{248}}$
We know that $\Delta {T_f}$ is directly proportional to the molarity. This relation can be written by the formula $\Delta {T_f} = {K_f}.m.i$
Substituting the above values, we get an equation $6 = 1.85 \times \dfrac{x}{{248}}$
By solving, we get $x = \dfrac{{6 \times 248}}{{1.85}}$
On further solving, we get $x = 804.38g$

Therefore, the required amount of ethylene glycol is $804.38g$.

Note: The relation between depression in freezing point and the molarity of the solute is given as $\Delta {T_f} = {K_f}.m.i$
Here, $\Delta {T_f}$ is the depression in freezing point, ${K_f}$ is the cryoscopic constant, m is the molarity and i is the van't hoff factor.