Question

(a) Calculate the vapor pressure of water above a solution prepared by adding \(22.5 \mathrm{~g}\) of lactose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) to \(200.0 \mathrm{~g}\) of water at \(338 \mathrm{~K}\). (Vapor-pressure data for water are given in Appendix B.) (b) Calculate the mass of propylene glycol \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{2}\right)\) that must be added to \(0.340 \mathrm{~kg}\) of water to reduce the vapor pressure by \(2.88\) torr at \(40^{\circ} \mathrm{C}\).

Short answer

The vapor pressure of water above the solution with lactose is 186.3 torr. To reduce the vapor pressure by 2.88 torr at 40°C, 79.1 g of propylene glycol must be added to the water.

Step-by-step solution

Find the mole fraction of water in the solution

Calculate the number of moles of both lactose and water in the solution to find their mole fractions. Given, mass of lactose = 22.5g and molar mass of lactose = 342.30 g/mol (12 Carbon, 22 Hydrogen, and 11 Oxygen) Number of moles of lactose = mass / molar mass = 22.5 g / 342.30 g/mol = \(6.57 \times 10^{-2} \) moles Also, mass of water = 200.0 g and molar mass of water, H2O = 18.015 g/mol Number of moles of water = mass / molar mass = 200.0 g / 18.015 g/mol = 11.11 moles Now find the mole fraction of water, X_water: X_water = moles of water / (moles of water + moles of lactose) = 11.11 / (11.11 + \(6.57 \times 10^{-2}\)) = 0.994

Apply Raoult's Law to find the vapor pressure of water above the solution

Raoult's Law relates the mole fraction of a solute with the vapor pressure of a solvent in a solution: P_solution = X_water * P0_water where P0_water is the vapor pressure of pure water at 338K. Refer to Appendix B to find the vapor pressure of pure water at 338K, which is 187.5 torr. Now calculate the vapor pressure of water above the solution: P_solution = 0.994 * 187.5 torr = 186.3 torr So, the vapor pressure of water above the solution with lactose is 186.3 torr. b) Mass of propylene glycol needed to reduce vapor pressure by 2.88 torr

Calculate the initial vapor pressure of water

The temperature is given as 40°C, which is equivalent to 313.15 K. Appendix B gives the vapor pressure of pure water at 313.15 K as 55.3 torr.

Determine the final vapor pressure of water

We are given that the vapor pressure should be reduced by 2.88 torr. Subtract this value from the initial vapor pressure: P_final = 55.3 torr - 2.88 torr = 52.42 torr

Calculate the mole fraction of water in the final solution

Using Raoult's Law, we can calculate the mole fraction of water in the final solution: X_water = P_final / P0_water = 52.42 torr / 55.3 torr = 0.948

Find the amount of propylene glycol needed

Given that the mass of water in the solution is 0.340 kg or 340 g, we can calculate the number of moles of water in the solution: Number of moles of water = mass / molar mass = 340 g / 18.015 g/mol = 18.88 moles Now we need to find the mole fraction of propylene glycol in the final solution. Since the sum of mole fractions must equal 1: X_propylene_glycol = 1 - X_water = 1 - 0.948 = 0.052 Then we calculate the number of moles of propylene glycol: Number of moles of propylene glycol = (moles of water * X_propylene_glycol) / X_water = (18.88 moles * 0.052) / 0.948 = 1.04 moles Finally, determine the mass of propylene glycol needed, using its molar mass, C3H8O2 = 76.094 g/mol: mass = moles * molar mass = 1.04 moles * 76.094 g/mol = 79.1 g So, 79.1 g of propylene glycol must be added to the water to reduce the vapor pressure by 2.88 torr at 40°C.