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?

To determine which beaker's solution level rises and which falls, we need to calculate the mole fraction of solute and solvent in each beaker. To do this, we must first find the moles of solute and solvent in each solution. For a nonvolatile nonelectrolyte: Concentration = 0.050 M Solvent: water, volume = 30 mL Number of moles of solute: \( 0.050 \mathrm{M} \times 0.030 \mathrm{L} = 0.0015 \mathrm{mol}\) Number of moles of solvent: \( \dfrac{0.030 \mathrm{L} }{18.015 \mathrm{\dfrac{g}{mol}} \times 1 \mathrm{\dfrac{g}{mL}} }\ = 1.666 \mathrm{mol}\) Mole fraction of solute: \( \dfrac{0.0015}{0.0015 + 1.666} = 0.0009\) For NaCl solution: Concentration = 0.035 M Solvent: water, volume = 30 mL Number of moles of solute: \( 0.035 \mathrm{M} \times 0.030 \mathrm{L} = 0.00105 \mathrm{mol}\) Number of moles of solvent: \( \dfrac{0.030 \mathrm{L} }{18.015 \mathrm{\dfrac{g}{mol}} \times 1 \mathrm{\dfrac{g}{mL}} }\ = 1.666 \mathrm{mol}\) Mole fraction of solute: \( \dfrac{0.00105}{0.00105 + 1.666} = 0.00063\) Now that we have the mole fraction of solute and solvent in each beaker, we can proceed to calculate the vapor pressure of each solution.

Using the mole fraction of solvent in each solution, we will now determine the vapor pressure of each solution using Raoult's Law: \( P_{solvent} = X_{solvent} \times P^*_\mathrm{solvent}\) Where \(P_{solvent}\) is the vapor pressure of the solvent in the solution, \(X_{solvent}\) is the mole fraction of the solvent, and \(P^*_\mathrm{solvent}\) is the vapor pressure of the pure solvent. The vapor pressure of pure water at \(25^{\circ} \mathrm{C}\) is 23.76 mmHg. For the nonvolatile nonelectrolyte solution: \( P_1 = X_{solvent\mathrm{(Nonelectrolyte)}} \times P^*_\mathrm{water}\) \( P_1 = (1 - 0.0009) \times 23.76 \mathrm{mmHg} = 23.58 \mathrm{mmHg}\) For the NaCl solution: \( P_2 = X_{solvent\mathrm{(NaCl)}} \times P^*_\mathrm{water}\) \( P_2 = (1 - 0.00063) \times 23.76 \mathrm{mmHg} = 23.65 \mathrm{mmHg}\) Now that we have the vapor pressure of each solution, we can determine which beaker's solution level rises and which falls.

Since the vapor pressure of the nonvolatile nonelectrolyte solution is lower than that of the NaCl solution, water vapor will move from the beaker with the NaCl solution to the beaker with the nonvolatile nonelectrolyte solution. As a result, the solution level in the beaker containing the nonvolatile nonelectrolyte will rise, and the solution level in the beaker containing the NaCl solution will fall.

The solution levels rise and fall until equilibrium is reached, i.e., when both solutions have the same vapor pressure. We can calculate the final volumes of the solutions in the two beakers using the mole fractions and vapor pressures at equilibrium. Let ΔV be the volume change in the NaCl solution beaker. \( X_{solvent_\mathrm{NaCl \mathrm{,final}}} = \dfrac{30-\Delta V}{30} \) Since the vapor pressures are equal at equilibrium, we can write: \( (1-X_{solvent_\mathrm{NaCl\mathrm{,final}}})P_{water}^* = (1-X_{solvent_\mathrm{Nonelectrolyte}})P_{water}^*\) Substituting the values and simplifying: \( (1 - \dfrac{30-\Delta V}{30})23.76 = (1 - 0.0009)23.76\) Solving for ΔV gives us: \( \Delta V = 0.025 \mathrm{L}\) So the final volume of the nonvolatile nonelectrolyte solution is 30 mL + 0.025 L = 30.025 mL, and the final volume of the NaCl solution is 30 mL - 0.025 L = 29.975 mL. In summary, the solution level rises in the beaker containing the nonvolatile nonelectrolyte, and it falls in the beaker containing the NaCl solution. The final volumes of the solutions in the two beakers when equilibrium is attained are 30.025 mL and 29.975 mL, respectively.