A solution containing $62g$ ethylene glycol in $250g$ water is cooled to $-{10}^{\circ }C.$ If $K_f$ for water is $1.86Kkgmo{l}^{-1}$, the amount of water $\left(ing\right)$ separated as ice is

The correct option is D 64
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.
$\text{Molality (m)}=\frac{\frac{w_{\text{solute}}}{M}\times 1000}{W_{\text{solvent}}}$
where,
$w$ is weight of solute
$W$ is weight of solvent
$M$ is molar mass of the solute

$\Rightarrow 10=1.86\times \frac{62\times 1000}{62\times W_{solvent}}$

$\Rightarrow W_{solvent}=\frac{1.86\times 1000}{10}$

$=186g$

Give amount of $H_{2}Ois250g$

$\therefore$ The amount of water separated as ice,
$=250-186=64g$