Calculate the amount of ethylene glycol to be added to $4kg$ of water to prevent it from freezing at $-6^{\circ }C$.
$(K_f\text{for water}=1.85Kkgmo{l}^{-1})$

The correct option is A $804.3g$
To prevent the water from freezing at $-6^{\circ }C$, the frezzing point of water should drop to $-6^{\circ }C$.
Thus, the depression in freezing point $△T_f=6^{\circ }C$
We know,
Freezing point depression of a solution is given by,
$△T_f=K_f\times m$
where,
$△T_f$ is the depression in freezing point.
$K_f$ is molal depression constant.
m is molality of solution.
Given:
$\Delta T=6^{\circ }C\text{Weight of solvent,}W=4kg=4000g\text{Molar mass of solute,}M=62g/mol{K}_f\text{of water}=1.85Kkgmo{l}^{-1}\text{Molality}=\frac{△T_f}{K_f}w=\frac{M\times W\times \Delta T}{1000\times K_f}$
where, w is the weight of the solute
$\therefore w=\frac{62\times 4000\times 6}{1000\times 1.85}=804.3g$

Thus, the amount of ethylene glycol to be added to $4kg$ of water to prevent it from freezing at $-6^{\circ }C$ is $804.3g$