Q. A solution containing 1.8 g of a compound (empirical formula $CH_{2}O$) in 40 g of water is observed to freeze at $- 0.465^{\circ}C.$ The molecular formula of the compound is ($K_{f}\ of\ water = 1.86\ kg\ K\ mol^{- 1}$)
A) $C_{2}H_{4}O_{2}$
B) $C_{3}H_{6}O_{3}$
C) $C_{4}H_{8}O_{4}$
D) $C_{6}H_{12}O_{6}$
Explanation
(d)
$M_{2} = \frac{K_{f} \times w_{2} \times 1000}{\bigtriangleup T_{f} \times w_{1}}\ \$
$or\ \ \ \ M_{2} = \frac{1.86 \times 1.8 \times 1000}{0.465 \times 40}$=180
n=$\frac{180}{emp.formula\ mass\ } = \ \frac{180}{30} = 6$
Molecular formula of the compound is $C_{6}H_{12}O_{6}$ .
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.