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<p>Understanding Raoult's Law & Vapor Pressure Lowering</p>
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9. 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 given data: molality (m) = 1.32 m, molecular weight (MW) of C\_6H\_12O\_6 = 180.156 g/mol, vapor pressure of pure water (P\_0) at 25°C = 23.8 torr.
Recall that vapor pressure lowering (\(\u\)0394P) is calculated using Raoult's Law for a non-volatile solute:
\(\$\Delta P = X_{solute} \times P_0\)
where \(X_{solute}\) is the mole fraction of the solute.
Calculate the mole fraction of the solute using the molality definition:
Molality (m) = moles of solute / kg of solvent. Assume 1 kg of solvent for simplicity, so moles of solute = 1.32 mol, moles of solvent (water) = 1000 g / 18.015 g/mol.
Calculate the mole fraction of the solute:
\(X_{solute} = \frac{\text{moles of solute}}{\text{moles of solute} + \text{moles of solvent}}\).
Use the mole fraction of the solute to find vapor pressure lowering:
\(\Delta P = X_{solute} \times P_0\). Then, calculate the vapor pressure of the solution as
\(P_{solution} = P_0 - \Delta P\).
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