Question

The freezing point of a solution containing $36\, g$ of a compound (empirical formula: $CH _{2} O$ ) in $1.20 \,kg$ of water is found to be $0.93^{\circ} C$. The molecular formula of the solute is $\left(K_{f}\right.$ of water is $\left.1.86 \,K Kgmol ^{-1}\right)$

• A. $C H_{2} O$
• B. $C _{3} H _{6} O _{3}$
• C. $C _{4} H _{8} O _{4}$
• D. $C _{2} H _{4} O _{2}$

Answer 1

Answer: (D)

$\Delta f=K_{f} \times m$
$0.93\, K=1.86 \times \frac{36 / M}{1.20}$
$M=\frac{1.86 \times 36}{0.93 \times 1.2}=60\, g\,mol ^{-1}$
$n=\frac{\text { Molar mass }}{\text { Empirical formula mass }}$
$n=\frac{60}{30}=2$
Hence molecular formula $=\left( CH _{2} O \right) \times 2= C _{2} H _{4} O _{2}$