91Ó°ÊÓ

To help us in applying Raoult's Law, we need to find the mole fractions of the solvents (water) in both solutions. For the nonvolatile nonelectrolyte solution: Molarity (\(C_1\)) = \(0.050 \mathrm{M}\) Volume (\(V_1\)) = \(30.0 \mathrm{~mL} = 0.030 \mathrm{L}\) Moles of solute 1 = \(C_1 * V_1 = 0.050 * 0.030 = 0.0015 \mathrm{mol}\) For the \(\mathrm{NaCl}\) solution: Molarity (\(C_2\)) = \(0.035 \mathrm{M}\) Volume (\(V_2\)) = \(30.0 \mathrm{~mL} = 0.030 \mathrm{L}\) Moles of solute 2 = \(C_2 * V_2 = 0.035 * 0.030 = 0.00105 \mathrm{mol}\) Now, we can find the moles of water (solvent) in both solutions. Moles of water in solution 1 = \(\frac{V_1}{18.015} = \frac{30}{18.015} = 1.6657 \mathrm{mol}\) Moles of water in solution 2 = \(\frac{V_2}{18.015} = \frac{30}{18.015} = 1.6657 \mathrm{mol}\) Calculating the mole fractions of water in both solutions, Mole fraction of water in solution 1 (\(X_{H_2O1}\)) = \(\frac{1.6657}{1.6657 + 0.0015} = 0.9991\) Mole fraction of water in solution 2 (\(X_{H_2O2}\)) = \(\frac{1.6657}{1.6657 + 0.00105} = 0.99937\)

Raoult's Law states that the partial pressure exerted by the solvent in a solution is directly proportional to its mole fraction. Partial pressure in solution 1 (\(P_{H_2O1}\)) = \(X_{H_2O1} * P^0_{H_2O}\) Partial pressure in solution 2 (\(P_{H_2O2}\)) = \(X_{H_2O2} * P^0_{H_2O}\) Here, \(P^0_{H_2O}\) is the vapor pressure of pure water, which is \(23.76 \mathrm{mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C}\). Now, we can calculate the partial pressure in both solutions: \(P_{H_2O1}\) = \(0.9991 * 23.76 = 23.7324 \mathrm{mm} \mathrm{Hg}\) \(P_{H_2O2}\) = \(0.99937 * 23.76 = 23.7389 \mathrm{mm} \mathrm{Hg}\)

Since the vapor pressure of solution 2 is higher than that of solution 1 (\(P_{H_2O2} > P_{H_2O1}\)), water will evaporate from solution 2 and condense in solution 1. This means the level of solution 1 will rise, and the level of solution 2 will fall. At equilibrium, the vapor pressure of both solutions will be equal, which means the moles of water that evaporate from solution 2 and condense in solution 1 will be equal. Let's denote this quantity as \(\Delta n_{H_2O}\). The new mole fractions of water in both solutions will be equal (\(X_{H_2O1}^{eq} = X_{H_2O2}^{eq}\)) and can be calculated as follows: \(X_{H_2O1}^{eq} = \frac{1.6657 + \Delta n_{H_2O}}{1.6657 + 0.0015 + \Delta n_{H_2O}}\) \(X_{H_2O2}^{eq} = \frac{1.6657 - \Delta n_{H_2O}}{1.6657 + 0.00105 - \Delta n_{H_2O}}\) We can now solve for \(\Delta n_{H_2O}\): \((1.6657 + \Delta n_{H_2O})(1.66675 - \Delta n_{H_2O}) = (1.66665 + \Delta n_{H_2O})(1.6675 - \Delta n_{H_2O})\) \(\Delta n_{H_2O} = 0.00025 \mathrm{mol}\) Finally, we can find the equilibrium volumes for both beakers: \(V_1^{eq} = (1.6657 + 0.00025) * 18.015 = 30.0799 \mathrm{mL}\) \(V_2^{eq} = (1.6657 - 0.00025) * 18.015 = 29.9201 \mathrm{mL}\) So, the solution level rises in beaker 1 and falls in beaker 2. The equilibrium volume of beaker 1 is approximately \(30.08 \mathrm{mL}\), while that of beaker 2 is approximately \(29.92 \mathrm{mL}\).