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

Answer: (C)

Verified step by step guidance

1. 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.
2. Recall that the vapor pressure lowering in a solution can be found using Raoult's Law for a non-volatile solute: \(P = X_{solvent} \times P_0\), where \(X_{solvent}\) is the mole fraction of the solvent (water) in the solution.
3. Calculate the mole fraction of the solvent. Start by assuming a convenient amount of solvent, typically 1 kg (1000 g) of water, to use molality definition: molality \(m = \frac{\text{moles of solute}}{\text{kg of solvent}}\). So, moles of solute = \(1.32 \times 1 = 1.32\) mol.
4. Calculate moles of solvent (water) using its molar mass (approximately 18.015 g/mol): \(\text{moles of water} = \frac{1000}{18.015}\). Then, find mole fraction of water: \(X_{solvent} = \frac{\text{moles of water}}{\text{moles of water} + \text{moles of solute}}\).
5. Finally, calculate the vapor pressure of the solution using Raoult's Law: \(P = X_{solvent} \times P_0\). This will give the vapor pressure of the solution at 25°C.