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?

Short answer

(a) The solution level in Beaker 1 will rise, and the solution level in Beaker 2 will fall. (b) The equilibrium volumes in Beaker 1 and Beaker 2 can be found using their initial volumes and the volume of water transferred between them: Equilibrium volume in Beaker 1 = Initial volume in Beaker 1 - x, Equilibrium volume in Beaker 2 = Initial volume in Beaker 2 + x, where x is calculated using the differences in vapor pressure and the overall volume of water in the sealed box.

Step-by-step solution

Calculate the mole fraction of water in both beakers.

First, we need to find the mole fraction of water in both solutions. To do this, we need the volume of water in each of the solutions, which is given, and the concentration of the solute in each solution, which is also given. For Beaker 1 (nonvolatile nonelectrolyte), volume = 30.0 mL Concentration = 0.050 M For Beaker 2 (NaCl solution), volume = 30.0 mL, Concentration = 0.035 M. We need to convert the volume to L and then multiply by the concentration to get the amount (mol) of solute in each beaker, and then find moles of water. Moles of water in Beaker 1 can be determined as: n_water_B1 = Volume of solution - mol of solute n_water_B1 = (30 mL / 1000)*(55.5 mol/L - 0.050 M) where 55.5 mol/L is the concentration of water. Similarly, moles of water in Beaker 2 can be determined as: n_water_B2 = (30 mL / 1000)*(55.5 mol/L - 0.035 M). Now we can find the mole fraction of water (X_water) in both beakers: X_water_B1 = n_water_B1 / (n_water_B1 + moles of solute) X_water_B2 = n_water_B2 / (n_water_B2 + moles of solute).

Apply Raoult's Law to calculate the final vapor pressure of each beaker.

According to Raoult's Law, the vapor pressure of each solution (P_soln) is the product of the mole fraction of water in the solution and the vapor pressure of pure water at the given temperature (P_water). We are given the temperature to be 25ºC. The vapor pressure of pure water at 25°C is approximately 23.76 mm Hg. Now we can calculate the vapor pressure for both beakers using this formula: P_soln_B1 = X_water_B1 * P_water P_soln_B2 = X_water_B2 * P_water

Compare the vapor pressures of both solutions.

Since the water vapor in the sealed box reached equilibrium, the total amount of water in both beakers must be equal. Compare the vapor pressures of the two solutions, which we calculated in Step 2. If P_soln_B1 > P_soln_B2, the solution level in Beaker 1 will fall, and the solution level in Beaker 2 will rise. Conversely, if P_soln_B1 < P_soln_B2, the solution level in Beaker 1 will rise, and the solution level in Beaker 2 will fall.

Calculate the volume of water transferred and find the equilibrium volumes.

Finally, to find the equilibrium volumes, we need to calculate the amount of water transferred between Beaker 1 and Beaker 2. We can do this using the differences in vapor pressure from Step 3 and the overall volume of water in the sealed box. Let x be the volume of water transferred from Beaker 1 to Beaker 2, then: (B1 initial volume - x)(P_soln_B1) = (B2 initial volume + x)(P_soln_B2). Solve for x. Now we can find the equilibrium volumes in both beakers: Equilibrium volume in Beaker 1 = Initial volume in Beaker 1 - x Equilibrium volume in Beaker 2 = Initial volume in Beaker 2 + x