Multiple Choice
Which of the following best describes what happens when water freezes?
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14. Solutions
Freezing Point Depression
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Which of the following aqueous solutions will have the lowest freezing point?
A
0.10 molal KNO_3
B
0.10 molal C_6H_{12}O_6 (glucose)
C
0.10 molal NaCl
D
0.10 molal CaCl_2
Verified step by step guidance1
Recall that the freezing point depression of a solution depends on the molality (m) of the solute and the van't Hoff factor (i), which accounts for the number of particles the solute dissociates into in solution. The formula for freezing point depression is: \[\Delta T_f = i \times K_f \times m\] where \(\Delta T_f\) is the freezing point depression, \(K_f\) is the freezing point depression constant of the solvent (water in this case), \(m\) is the molality, and \(i\) is the van't Hoff factor.
Determine the van't Hoff factor (\(i\)) for each solute: - For glucose (\(C_6H_{12}O_6\)), a molecular compound that does not dissociate, \(i = 1\). - For \(KNO_3\), which dissociates into \(K^+\) and \(NO_3^-\) ions, \(i = 2\). - For \(NaCl\), which dissociates into \(Na^+\) and \(Cl^-\) ions, \(i = 2\). - For \(CaCl_2\), which dissociates into \(Ca^{2+}\) and 2 \(Cl^-\) ions, \(i = 3\).
Since all solutions have the same molality (0.10 molal), the freezing point depression depends directly on the van't Hoff factor \(i\). Calculate the effective concentration of particles for each: - Glucose: \$0.10 \times 1 = 0.10\( - \)KNO_3\(: \)0.10 \times 2 = 0.20\( - \)NaCl\(: \)0.10 \times 2 = 0.20\( - \)CaCl_2\(: \)0.10 \times 3 = 0.30$
Recognize that the solution with the highest effective particle concentration (highest \(i \times m\)) will have the greatest freezing point depression, thus the lowest freezing point.
Conclude that among the given options, the 0.10 molal \(CaCl_2\) solution has the highest van't Hoff factor and therefore will have the lowest freezing point.
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