Let's break down the question and verify the answer: **Given:** - Mass of sucrose (C₁₂H₂₂O₁₁) = **158.0 g** - Molar mass of sucrose = **342.3 g/mol** - Volume of water = **643.5 cm³** - Density of water at 25°C = **0.9971 g/cm³** - Vapor pressure of pure water at 25°C = **23.76 torr** ### Step 1: Calculate the Moles of Sucrose \[ \text{Moles of sucrose} = \frac{158.0 \text{ g}}{342.3 \text{ g/mol}} = 0.4617 \text{ mol} \] ### Step 2: Calculate the Mass and Moles of Water \[ \text{Mass of water} = 643.5 \text{ cm}^3 \times 0.9971 \text{ g/cm}^3 = 641.65 \text{ g} \] \[ \text{Moles of water} = \frac{641.65 \text{ g}}{18.016 \text{ g/mol}} = 35.63 \text{ mol} \] ### Step 3: Calculate Mole Fraction of Water \[ \text{Total moles} = 0.4617 + 35.63 = 36.09 \] \[ \text{Mole fraction of water} = \frac{35.63}{36.09} = 0.9873 \] ### Step 4: Use Raoult's Law to Find Vapor Pressure of Solution \[ P_{\text{solution}} = X_{\text{water}} \times P^0_{\text{water}} \] \[ P_{\text{solution}} = 0.9873 \times 23.76 \text{ torr} = 23.46 \text{ torr} \] ### Step 5: Check the Options But the answer marked as correct is **23.86 torr** (option a), but our calculation gives **23.46 torr** (option d). #### Let's check for possible rounding or small calculation errors: - 0.4617/36.09 ≈ 0.0128 (sucrose fraction) - 35.63/36.09 ≈ 0.9872 (water fraction) \[ 0.9872 \times 23.76 = 23.46 \text{ torr} \] ### **Conclusion** - The answer based on calculation is **23.46 torr** (option d). - However, the answer marked is **23.86 torr** (option a), which may be a typo or calculation error in the options. **Calculation details show that the correct answer should be 23.46 torr (d), not 23.86 torr (a).** If you want to discuss or clarify further, let me know!