How many grams of glucose, C₆H₁₂O₆, must be added to 515.0 g of water to give a solution with a vapor pressure of 13.2 torr at 20.0ºC? The vapor pressure of pure water at 20.0ºC is 17.5 torr.

• A. 9.54 × 10² g
• B. 1.68 × 10³ g
• C. 5.29 g
• D. 9.31 g

Answer: (B)

Verified step by step guidance

1. Identify the known values: the vapor pressure of pure water (P₀) is 17.5 torr, the vapor pressure of the solution (P) is 13.2 torr, and the mass of water is 515.0 g.
2. Use Raoult's Law to find the mole fraction of water in the solution: P = X_water * P₀, where X_water is the mole fraction of water.
3. Rearrange Raoult's Law to solve for the mole fraction of water: X_water = P / P₀.
4. Calculate the moles of water using its mass and molar mass (18.02 g/mol): moles of water = mass of water / molar mass of water.
5. Use the mole fraction of water to find the moles of glucose: X_water = moles of water / (moles of water + moles of glucose). Rearrange to solve for moles of glucose, then convert moles of glucose to grams using its molar mass (180.18 g/mol).