Multiple Choice
Which of the following best describes the effect of adding a nonvolatile solute to a solvent on its boiling and freezing points?
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14. Solutions
Boiling Point Elevation
Multiple Choice
Which of the following aqueous solutions is most likely to have the highest boiling point?
A
1.0 M C6H12O6 (glucose)
B
1.0 M NaCl
C
1.0 M KBr
D
1.0 M CaCl2
Verified step by step guidance1
Understand that the boiling point elevation of a solution depends on the number of solute particles dissolved in the solvent, not just the concentration in molarity. This is a colligative property, which depends on the total concentration of dissolved particles.
Recall the boiling point elevation formula: \(\Delta T_b = i \cdot K_b \cdot m\), where \(\Delta T_b\) is the boiling point elevation, \(i\) is the van't Hoff factor (number of particles the solute dissociates into), \(K_b\) is the boiling point elevation constant of the solvent, and \(m\) is the molality of the solution.
Determine the van't Hoff factor \(i\) for each solute: glucose (\(C_6H_{12}O_6\)) does not dissociate, so \(i=1\); NaCl dissociates into Na\(^+\) and Cl\(^-\), so \(i=2\); KBr dissociates into K\(^+\) and Br\(^-\), so \(i=2\); CaCl\(_2\) dissociates into Ca\(^{2+}\) and 2 Cl\(^-\) ions, so \(i=3\).
Since all solutions are 1.0 M, compare their effective particle concentrations by multiplying molarity by \(i\): glucose (1.0 x 1 = 1), NaCl (1.0 x 2 = 2), KBr (1.0 x 2 = 2), CaCl\(_2\) (1.0 x 3 = 3). The higher the effective particle concentration, the greater the boiling point elevation.
Conclude that the 1.0 M CaCl\(_2\) solution, having the highest van't Hoff factor and thus the greatest number of dissolved particles, will have the highest boiling point among the given options.
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