The empirical formula of a non-electrolyte is \(\mathrm{CH}_{2} \mathrm{O}\). A solution containing \(3 \mathrm{~g}\) of the compound exerts the same osmotic pressure as that of \(0.05 \mathrm{M}\) glucose solution. The molecular formula of the compound is

For isotonic solution (solution with same osmotic pressure) \(C_{1}=C_{2}\)
\[
\begin{gathered}
\frac{W_{x}}{M_{X}}=\frac{W_{\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}}{M_{\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}} \\
\frac{3}{M_{x}}=\frac{9}{180} \\
0.05 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \text { containing } 9 \mathrm{~g} \text { of glucose } \\
M_{x}=\frac{3 \times 180}{9}=60
\end{gathered}
\]
Molecular weight of a non-electrolyte is 60
Molecular formula \(=n \times\) empirical formula
Empirical formula \(=\mathrm{CH}_{2} \mathrm{O}\)
Empirical formula mass \(=12+2+16=30\)
\[
\begin{aligned}
n &=\frac{\text { molecular mass }}{\text { empirical formula mass }} \\
&=\frac{60}{30}=2
\end{aligned}
\]
Molecular formula \(=2 \times \mathrm{CH}_{2} \mathrm{O}=\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\)