Multiple Choice
Which of the following substances would you predict to have the highest enthalpy of vaporization (ΔH_{vap})?
61
views
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
Start by understanding the concept of vapor pressure lowering, which is a colligative property. It depends on the number of solute particles in a solution, not their identity.
Use Raoult's Law for vapor pressure lowering: ΔP = X_solute * P°_solvent, where ΔP is the vapor pressure lowering, X_solute is the mole fraction of the solute, and P°_solvent is the vapor pressure of the pure solvent.
Calculate the mole fraction of the solute (C6H12O6). First, find the moles of solute using the molality (m = 1.32 mol/kg) and assume 1 kg of water as the solvent. Moles of solute = 1.32 mol.
Calculate the moles of water. The molar mass of water is approximately 18.015 g/mol, so 1 kg of water is 1000 g, which is 1000 g / 18.015 g/mol = 55.51 mol.
Determine the mole fraction of the solute: X_solute = moles of solute / (moles of solute + moles of solvent). Substitute the values to find X_solute, then use Raoult's Law to calculate ΔP = X_solute * 23.8 torr.
Related Videos
Related Practice
