(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 \(384 \mathrm{~Pa}\) at \(40^{\circ} \mathrm{C}\).

First, we need to find the number of moles of lactose in 22.5 g. The molar mass of lactose (C12H22O11) is (12*12 + 22 + 16*11) g/mol = 342 g/mol. Number of moles of lactose = (Mass of lactose) / (Molar mass of lactose) Number of moles of lactose = 22.5 / 342 Number of moles of lactose = 0.0658 moles Molality (m) is the number of moles of solute per kilogram of solvent. In this case, the solvent is 200.0 g of water which is equal to 0.200 kg. Molality = (0.0658 moles of lactose) / (0.200 kg of water) Molality = 0.329 mol/kg

Raoult's law states that the vapor pressure of a solvent above a solution is directly proportional to the mole fraction of the solvent in the solution. The equation is: Vapor Pressure of the Solution = (Mole Fraction of Solvent) * (Vapor Pressure of Pure Solvent) Mole fraction of solvent (X_solvent) = (Molality of the Solution) / (Molality of the Solution + 1) In this case: X_solvent = 0.329 / (0.329 + 1) X_solvent = 0.329 / 1.329 X_solvent = 0.247 From Appendix B, vapor pressure of pure water at 338 K is 187.5 kPa. Vapor Pressure of the Solution = (0.247) * (187.5 kPa) Vapor Pressure of the Solution = 46.3 kPa (approximately)

The vapor pressure of water above the lactose solution is approximately 46.3 kPa.

In part (b), we need to calculate the mass of propylene glycol (C3H8O2) needed to be added to 0.340 kg of water to reduce the vapor pressure by 384 Pa. First, we need to find the vapor pressure of water at 40°C (313 K). From Appendix B, the vapor pressure of water at 313 K is 7.36 kPa. Let's denote the mass of propylene glycol to be added as 'm' grams. The molar mass of propylene glycol is (3*12 + 8 + 2*16) g/mol = 76 g/mol. Moles of propylene glycol = m / 76 Molality (m_prop_glycol) = Moles of propylene glycol / 0.340 kg_solvent We know that lowering the vapor pressure can be calculated using the formula: Vapor Pressure change = Molality * (Molar Mass of solvent) * constant Where the constant is given for water in the problem as 384 Pa. 7.36 kPa - (0.001 * molality * 18.015) kPa = 7.36 kPa - 384 Pa Since 1 kPa = 1000 Pa, we can write: 7.36 - (0.001 * molality * 18.015) = 7.36 - 384/1000 7.36 - (0.001 * molality * 18.015) = 6.976 Solving for molality: 0.001 * molality * 18.015 = 0.384 molality * 18.015 = 384 molality = 384 / 18.015 molality = 21.34 We know that molality (m_prop_glycol) = Moles of propylene glycol / 0.340 kg_solvent, so: Moles of propylene glycol = 21.34 * 0.340 Moles of propylene glycol = 7.25 Finally, we can find the mass of propylene glycol needed: Mass of propylene glycol = Moles of propylene glycol * Molar Mass of propylene glycol Mass of propylene glycol = 7.25 * 76 Mass of propylene glycol = 550.9 g

The mass of propylene glycol that must be added to 0.340 kg of water to reduce the vapor pressure by 384 Pa at 40°C is approximately 550.9 g.