A quantity of 3.125 g of a mixture of KCl and NaCl dissolved in 1 kg of water produces a depression — Solutions and Colligative Properties Chemistry Question
Question
A quantity of 3.125 g of a mixture of KCl and NaCl dissolved in 1 kg of water produces a depression of $0.186^\circ\text{C}$ in freezing point. The molar ratio of KCl to NaCl in the solution (assuming complete dissociation of the salts) is ($K_f = 1.86 \text{ deg/molal}$)
💡 Solution & Explanation
Both KCl and NaCl yield 2 ions ($i=2$). $\Delta T_f = i \times K_f \times (n_{KCl} + n_{NaCl}) \Rightarrow 0.186 = 2 \times 1.86 \times (n_{KCl} + n_{NaCl})$. Thus, $n_{KCl} + n_{NaCl} = 0.05 \text{ mol}$. Let $n_{KCl} = x$ and $n_{NaCl} = 0.05 - x$. The mass equation is $74.5x + 58.5(0.05 - x) = 3.125$. Expanding gives $74.5x + 2.925 - 58.5x = 3.125 \Rightarrow 16x = 0.2 \Rightarrow x = 0.0125 \text{ mol}$. Then, $n_{NaCl} = 0.05 - 0.0125 = 0.0375 \text{ mol}$. The molar ratio is $0.0125 / 0.0375 = 1/3$ or 1:3. Therefore, correct answer is A.