The freezing point of aqueous solution that contains $3$% urea, $7.45$% $KCl$ and $9$% of glucose is: (given $K_1$ of water $=1.86$ and assume $molarity=molality$).

The correct option is B $267.42K$
urea = $3\%$

KCl = $7.45\%$

Glucose = $9\%$

Let Amount of water be $100gmN{H}_{2}CON{H}_2$ urea = $3gm$ molar mass = $60$

$KCl=7.45g$ molar mass = $74.5$ $C_6{H}_{12}{O}_6$ Glucose = $9$g Molar mass = $180$

total no of mole = $\frac{3}{60}+\frac{7.45}{74.5}+\frac{9}{180}$

$\frac{1}{20}+\frac{1}{10}+\frac{1}{20}$

$=\frac{1}{5}=0.2$

molality = $\frac{0.2}{100}=2m$

$\Delta T_f=k_{f}m$

$=1.86\times 2=3.72$

Freezing point = $273-3.72$

$=269.28k$