Based on the data given in Table 13.4, which solution would give ...

Question

Based on the data given in Table 13.4, which solution would give the larger freezing-point lowering, a 0.030 m solution of NaCl or a 0.020 m solution of K2SO4?

Accepted Answer

Step 1: Understand the concept of freezing-point depression, which is a colligative property. It depends on the number of solute particles in a solution, not their identity.. Step 2: Use the formula for freezing-point depression: $ \Delta T_f = i \cdot K_f \cdot m $, where $ \Delta T_f $ is the freezing-point depression, $ i $ is the van't Hoff factor, $ K_f $ is the freezing-point depression constant, and $ m $ is the molality of the solution.. Step 3: Determine the van't Hoff factor $ i $ for each solute. For NaCl, which dissociates into Na$^+$ and Cl$^-$, $ i = 2 $. For K$_2$SO$_4$, which dissociates into 2 K$^+$ and SO$_4$$^{2-}$, $ i = 3 $.. Step 4: Calculate the effective molality for each solution by multiplying the molality by the van't Hoff factor: $ 0.030 \text{ m} \times 2 $ for NaCl and $ 0.020 \text{ m} \times 3 $ for K$_2$SO$_4$.. Step 5: Compare the effective molalities to determine which solution has a larger freezing-point lowering. The solution with the higher effective molality will have a greater freezing-point depression.