When 2 g non-volatile hydrocarbon containing 94.4% carbon by mass, is dissolved in 100 g benzene, th — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

The hydrocarbon has 94.4% C and 5.6% H. Moles of C $= 94.4 / 12 = 7.86$. Moles of H $= 5.6 / 1 = 5.6$. Ratio is $7.86 : 5.6 \approx 1.4 : 1 = 7 : 5$. The empirical formula is $C_7H_5$ (Empirical mass = 89). Using Raoult's law: $(P^\circ - P) / P = n / N$. $(89.78 - 89.0) / 89.0 = (2/M) / (100/78)$. $0.78 / 89.0 = (2/M) \times 0.78 \Rightarrow 1/89.0 = 2/M \Rightarrow M = 178$ g/mol. Since $178 / 89 = 2$, the molecular formula is $(C_7H_5)_2 = C_{14}H_{10}$. Therefore, correct answer is D.