Multiple Choice
Which of the following substances would you predict to have the highest enthalpy of vaporization (ΔH_{vap})?
116
views
14. Solutions
Vapor Pressure Lowering (Raoult's Law)
Multiple Choice
Determine the vapor pressure 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 key concept: Vapor pressure lowering is a colligative property that depends on the mole fraction of the solute in the solution. The presence of a non-volatile solute lowers the vapor pressure of the solvent.
Calculate the mole fraction of the solute. Since the molality (m) is given as 1.32 m, this means 1.32 moles of solute per 1 kg of solvent (water). Use this to find the moles of solvent: moles of solvent = mass of solvent (kg) / molar mass of solvent (g/mol). For water, molar mass is approximately 18.015 g/mol.
Calculate the mole fraction of the solvent (\(X_{solvent}\)) using the formula:
\[X_{solvent} = \frac{n_{solvent}}{n_{solvent} + n_{solute}}\]
where \(n_{solute}\) is the moles of solute (given by molality) and \(n_{solvent}\) is the moles of solvent calculated in the previous step.
Use Raoult's Law to find the vapor pressure of the solution:
\[P_{solution} = X_{solvent} \times P^0_{solvent}\]
where \(P^0_{solvent}\) is the vapor pressure of pure water (23.8 torr).
Calculate the vapor pressure lowering (\(\Delta P\)) by subtracting the vapor pressure of the solution from the vapor pressure of pure solvent:
\[\Delta P = P^0_{solvent} - P_{solution}\]
This value represents how much the vapor pressure is lowered due to the solute.
Related Videos
Related Practice
