When $36 \mathrm{~g}$ of a non-volatile, non-electrolytic solute having the empirical formula $\mathrm{CH}_2…
When $36 \mathrm{~g}$ of a non-volatile, non-electrolytic solute having the empirical formula $\mathrm{CH}_2 \mathrm{O}$ is dissolved in $1.2 \mathrm{~kg}$ of water, the solution freezes at $-0.93^{\circ} \mathrm{C}$. The molecular formula of the solute is $\left(K_f\right.$ of water $\left.=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$
$\mathrm{CH}_2 \mathrm{O}$
$\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2$
$\mathrm{C}_3 \mathrm{H}_6 \mathrm{O}_3$
$\mathrm{C}_4 \mathrm{H}_8 \mathrm{O}_4$
Solution
$
\begin{aligned}
\Delta t & =i K_f m \\
\Delta t & =\text { Depression in freezing point } \\
& =\left(t_1-t_2\right)=(0-(-0.93)) \\
\Delta t & =+0.93^{\circ} \mathrm{C}
\end{aligned}
$
$
\begin{aligned}
& i=\text { for non-electrolytic solute }=1 \\
& K_f=\text { constant }=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \\
& m=\text { molarity }=\frac{\text { moles of solute }}{\mathrm{kg} \text { of solvent }} \\
& =\frac{\text { weight in of solute }}{\text { molecular weight of solute } \times \mathrm{kg} \text { of solvent }} \\
& =\frac{36}{x \times 1.2} \\
&
\end{aligned}
$
$
\begin{aligned}
\therefore \quad 0.93 & =1 \times 1.86 \times \frac{36}{x \times 1.2} \\
x & =\frac{1.86 \times 36}{1.2 \times 0.93}=\frac{66.96}{1.116}=60 \\
\mathrm{C} & =12, \mathrm{H}=1, \mathrm{O}=16
\end{aligned}
$
For $\mathrm{CH}_2 \mathrm{O}=12+2+16=30$
$
\frac{60}{30}=2
$
$\therefore$ Formula of solute $=\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2$