When 36.0 g of a solute having the empirical formula is dissolved in 1.20 kg of water, the solution — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

$\Delta T_f = 0.93^\circ\text{C}$. From $\Delta T_f = K_f \times m$, $0.93 = 1.86 \times (36.0 / (M \times 1.20))$. $0.93 = 1.86 \times (30 / M) = 55.8 / M \Rightarrow M = 55.8 / 0.93 = 60$ g/mol. The empirical formula $CH_2O$ has a mass of $12 + 2(1) + 16 = 30$ g/mol. The multiple $n = M / M_{\text{empirical}} = 60 / 30 = 2$. Multiplying the empirical formula by 2 gives $C_2H_4O_2$. Therefore, correct answer is D.