A compound X has empirical formula $C X_{3} H X_{6} O$ (empirical formula mass $58 g mo l^{- 1}$ ). When $0.87 g$ of X is dissolved in $25.0 g$ of benzene ( $K_{f} = 5.12 K kg mo l^{- 1}$ ), the freezing point of benzene falls by $1.02 K$ . Assuming X neither associates nor dissociates in benzene, its molecular formula is

Answer: B

M = \frac{1000 K _{f} w}{Δ T _{f} W} = \frac{1000 \times 5.12 \times 0.87}{1.02 \times 25.0} = 174.7 g mo l^{- 1}

. The ratio to the empirical mass is

174.7/58 = 3.01 \approx 3

, so the molecular formula is

3 \times C X_{3} H X_{6} O = C X_{9} H X_{18} O X_{3}