The freezing point of a solution containing 36 g of a compound (empirical formula: ${\mathrm{CH}}_2\mathrm{O}$) in 1.20 kg of water is found to be - 0.93 °C. The molecular formula of the solute is

$-\Delta T_{\mathrm{f}}=K_{\mathrm{f}}m\Rightarrow 0.93 \mathrm{K}=\left(1.86 \mathrm{K} \mathrm{kg}{ \mathrm{mol}}^{-1}\right)\left(\frac{36 \mathrm{g}/M}{1.20 \mathrm{kg}}\right)$

This gives $M=\frac{1.86\times 36}{0.93\times 1.20} \mathrm{g}{ \mathrm{mol}}^{-1}=60 \mathrm{g}{ \mathrm{mol}}^{-1}$

$n=\frac{\text{ Molar mass }}{\text{ Molar Empirical formula mass }}=\frac{60 \mathrm{g}{ \mathrm{mol}}^{-1}}{30 \mathrm{g}{ \mathrm{mol}}^{-1}}=2$

Hence, molecular formula is ${\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{O}}_2$