The empirical formula of a non-electrolyte is ${\mathrm{CH}}_2\mathrm{O}.$ A solution containing $6\mathrm{g}$ of the compound exerts the same osmotic pressure as that of $0.05\mathrm{M}$ glucose solution at the same temperature. The molecular formula of the compound is 

For isotonic solution : $\frac{w_1}{m_1{V}_1}=\frac{w_2}{m_2{V}_2}$

$w_1=$mass of glucose$=0.05\times 180\mathrm{g}=9\mathrm{g}$

$m_1\mathit{=}$molecular mass of glucose$=180\mathrm{g}$

Assuming $V_1=V_2=1L$

$w_2=$mass of compound $=6g$

$m_2\mathit{=}$ molecular mass of compound = ?

$\begin{array}{l}\therefore \frac{9}{180}=\frac{6}{x}\Rightarrow x=120\mathrm{g}\\ \therefore \frac{\text{ Molecular mass }}{\text{ Empirical mass }}=n\Rightarrow n=\frac{120}{30}=4\end{array}$

$\therefore$Molecular formula $={\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_4\text{. }$