Problem 95 - Chemistry: The Central Science

Question

Two beakers are placed in a sealed box at \(25^{\circ} \mathrm{C}\). One beaker contains \(30.0 \mathrm{~mL}\) of a \(0.050 \mathrm{M}\) aqueous solution of a nonvolatile nonelectrolyte. The other beaker contains \(30.0 \mathrm{~mL}\) of a \(0.035 \mathrm{M}\) aqueous solution of \(\mathrm{NaCl}\). The water vapor from the two solutions reaches equilibrium. (a) In which beaker does the solution level rise, and in which one does it fall? (b) What are the volumes in the two beakers when equilibrium is attained, assuming ideal behavior?

Step-by-Step Solution

Answer

The solution level rises in the beaker containing the nonvolatile nonelectrolyte solution and falls in the beaker containing the NaCl solution. At equilibrium, the volume in the beaker with the nonvolatile nonelectrolyte solution is approximately \(30.08 \mathrm{mL}\), while the volume in the beaker with the NaCl solution is approximately \(29.92 \mathrm{mL}\).

Step 1: Calculate mole fractions of the solvents in both solutions.

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 (\(X_{H_2O2}\)) = \(\frac{1.6657}{1.6657 + 0.00105} = 0.99937\)

Step 2: Apply Raoult's Law

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: \(P_{H_2O1}\) = \(0.9991 * 23.76 = 23.7324 \mathrm{mm} \mathrm{Hg}\) \(P_{H_2O2}\) = \(0.99937 * 23.76 = 23.7389 \mathrm{mm} \mathrm{Hg}\)

Step 3: Determine the change in solution levels and equilibrium volumes

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 \(29.92 \mathrm{mL}\).