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

Answer: (C)

Verified step by step guidance

1. 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.
2. 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.
3. 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.
4. 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.
5. 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.