Question

The empirical formula of a non-electrolyte is $CH_2O$. A solution containing $6 \,g$ of the compound exerts the same osmotic pressure as that of $0.05\, M$ glucose solution at the same temperature. The molecular formula of the compound is

• A. $CH _{2} O$
• B. $C _{2} H_{4} O _{2}$
• C. $C_4H_8O_4$
• D. $C_3H_{6}O_3$

Answer 1

Answer: (B)

$C _{2} H_{4} O _{2}$

Answer 2

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