6 g of a compound exerts the same osmotic pressure as that of 0.05 M glucose solution. Find out the molecular formula of the compound if empirical formula of non-electrolyte is CH₂O.

Osmotic pressure $\propto$ concentration
$\therefore$ Concentration of the compound = Concentration of the glucose
       $$\begin{gathered} \frac{6}{\mathrm{M}}=0.05 \\ \mathrm{M}=120 \end{gathered}$$
Empirical formula (CH₂O) mass = 12+ 2+ 16 = 30 g mol^-1
          $$\begin{gathered} \mathrm{n}=\frac{\mathrm{molecular} \mathrm{mass}}{\mathrm{empirical} \mathrm{formula} \mathrm{mass}} \\ =\frac{120}{36}=4 \end{gathered}$$
Molecular formula $={\left({\mathrm{CH}}_2\mathrm{O}\right)}_{\mathrm{n}}={\left({\mathrm{CH}}_2\mathrm{O}\right)}_4={\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_4$