When 36   g of a non-volatile, non-electrolytic solute have the empirical formula CH 2 O is dissolved…

Given,

Mass of the solute $=36 \mathrm{g}$

Mass of the solvent $=1.2 \mathrm{kg}$

$∆{\mathrm{T}}_{\mathrm{f}}=$ Depression in freezing point $=0.93°\mathrm{C}$

${\mathrm{K}}_{\mathrm{f}}\left({\mathrm{H}}_2\mathrm{O}\right)=1.86 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}$

We know that,

$∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\times \mathrm{m}$

Molality, $\mathrm{m}=\frac{\mathrm{Mass} \mathrm{of} \mathrm{solute}}{\mathrm{Molar} \mathrm{mass} \mathrm{of} \mathrm{solute}\times \mathrm{mass} \mathrm{of} \mathrm{solvent} \left(\mathrm{kg}\right)}$

$\Rightarrow \mathrm{m}=\frac{36}{\mathrm{M}\times 1.2}$

Substituting the values, we get,

$$\begin{gathered} \Rightarrow 0.93=\frac{1.86\times 36}{\mathrm{M}\times 1.2} \\ \Rightarrow \mathrm{M}=60 \end{gathered}$$

Given,

Empirical formula $={\mathrm{CH}}_2\mathrm{O}$

Empirical Mass $=30$

So, 

$\mathrm{n}=\frac{60}{30}=2$

Hence, the molecular formula  $=2\left({\mathrm{CH}}_2\mathrm{O}\right)={\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{O}}_2$