Multiple Choice
Which of the following substances would you predict to have the highest enthalpy of vaporization (ΔH_{vap})?
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14. Solutions
Vapor Pressure Lowering (Raoult's Law)
Multiple Choice
Determine the vapor pressure lowering associated with 1.32 m C6H12O6 solution (MW:180.156 g/mol) at 25°C. The vapor pressure of pure water at 25°C is 23.8 torr.
A
0.553 torr
B
27.6 torr
C
23.2 torr
D
0.976 torr
Verified step by step guidance1
Identify the colligative property involved: vapor pressure lowering, which depends on the mole fraction of the solute in the solution.
Recall Raoult's Law for vapor pressure lowering: \(\Delta P = X_{solute} \times P^0_{solvent}\), where \(\Delta P\) is the decrease in vapor pressure, \(X_{solute}\) is the mole fraction of the solute, and \(P^0_{solvent}\) is the vapor pressure of the pure solvent.
Calculate the mole fraction of the solute using the molality (m) given. Since molality is moles of solute per kilogram of solvent, assume 1 kg of solvent for simplicity. Calculate moles of solute: \(n_{solute} = 1.32\) mol (since molality is 1.32 mol/kg). Calculate moles of solvent: \(n_{solvent} = \frac{1000 \text{ g}}{18.015 \text{ g/mol}}\) (molar mass of water). Then, mole fraction of solute is \(X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}}\).
Use the mole fraction of solute to find the vapor pressure lowering: \(\Delta P = X_{solute} \times P^0_{solvent}\), where \(P^0_{solvent} = 23.8\) torr.
Subtract the vapor pressure lowering from the pure solvent vapor pressure to find the vapor pressure of the solution: \(P_{solution} = P^0_{solvent} - \Delta P\).
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