Multiple Choice
Which of the following is closest to the freezing point of argon in kelvin?
14. Solutions
Freezing Point Depression
Multiple Choice
Which of the following aqueous solutions has the lowest freezing point?
A
0.10 molal CaCl_2
B
0.10 molal KNO_3
C
0.10 molal C_6H_{12}O_6 (glucose)
D
0.10 molal NaCl
Verified step by step guidance1
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. 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, 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 3 ions (\(\mathrm{Ca^{2+}}\) and 2 \(\mathrm{Cl^-}\)), so \(i = 3\).
- For \(\mathrm{KNO_3}\), it dissociates into 2 ions (\(\mathrm{K^+}\) and \(\mathrm{NO_3^-}\)), so \(i = 2\).
- For glucose (\(\mathrm{C_6H_{12}O_6}\)), it does not dissociate, so \(i = 1\).
- For \(\mathrm{NaCl}\), it dissociates into 2 ions (\(\mathrm{Na^+}\) and \(\mathrm{Cl^-}\)), so \(i = 2\).
Since all solutions have the same molality (\$0.10\( molal) and are in the same solvent (water), the freezing point depression depends directly on the van't Hoff factor \)i$. Calculate the effective concentration of particles as \(i \times m\) for each solution.
Compare the values of \(i \times m\) for each solution to determine which has the greatest freezing point depression (largest \(\Delta T_f\)), which corresponds to the lowest freezing point.
Conclude that the solution with the highest \(i \times m\) value will have the lowest freezing point. Therefore, the solution with \(\mathrm{CaCl_2}\), having \(i = 3\), will have the greatest freezing point depression and thus the lowest freezing point.
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