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Multiple Choice
Which of the following aqueous solutions would have the lowest freezing point?
A
0.10 molal CaCl_2
B
0.10 molal NaCl
C
0.10 molal C_6H_{12}O_6 (glucose)
D
0.10 molal KNO_3
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Verified step by step guidance
1
Recall that the freezing point depression of a solution depends on the molality of the solute and the number of particles the solute dissociates into in solution. The formula for freezing point depression is \(\Delta T_f = i \cdot K_f \cdot m\), where \(\Delta T_f\) is the freezing point depression, \(i\) is the van't Hoff factor (number of particles the solute dissociates into), \(K_f\) is the freezing point depression constant of the solvent (water in this case), and \(m\) is the molality of the solution.
Identify the van't Hoff factor \(i\) for each solute:
- For \(\mathrm{CaCl_2}\), it dissociates into 1 \(\mathrm{Ca^{2+}}\) and 2 \(\mathrm{Cl^-}\) ions, so \(i = 3\).
- For \(\mathrm{NaCl}\), it dissociates into 1 \(\mathrm{Na^+}\) and 1 \(\mathrm{Cl^-}\) ion, so \(i = 2\).
- For glucose (\(\mathrm{C_6H_{12}O_6}\)), it does not dissociate, so \(i = 1\).
- For \(\mathrm{KNO_3}\), it dissociates into 1 \(\mathrm{K^+}\) and 1 \(\mathrm{NO_3^-}\) ion, so \(i = 2\).
Since all solutions have the same molality (\$0.10$ molal), compare the products \(i \cdot m\) for each solution to determine which will have the greatest freezing point depression. The solution with the highest \(i \cdot m\) will have the lowest freezing point.
Calculate the effective molality (or total particle concentration) for each solution by multiplying the molality by the van't Hoff factor:
- \(\mathrm{CaCl_2}\): \(3 \times 0.10 = 0.30\)
- \(\mathrm{NaCl}\): \(2 \times 0.10 = 0.20\)
- Glucose: \(1 \times 0.10 = 0.10\)
- \(\mathrm{KNO_3}\): \(2 \times 0.10 = 0.20\)
Conclude that the solution with the highest effective molality (\$0.30$ for \(\mathrm{CaCl_2}\)) will have the greatest freezing point depression and thus the lowest freezing point.