Part A) If 0.220 mol of a nonvolatile nonelectrolyte are dissolved in 3.90 mol of water, what is the vapor pressure [tex]P_{H_2O}[/tex] of the resulting solution? The vapor pressure of pure water is 23.8 torr at 25°C.

Part B) A solution is composed of 1.00 mol cyclohexane $(P_{\text{cy}}^{\circ} = 97.6 \text{ torr})$ and 2.40 mol acetone $(P_{\text{ac}}^{\circ} = 229.5 \text{ torr})$. What is the total vapor pressure [tex]P_{\text{total}}[/tex] above this solution?

In Part A, the vapor pressure of the resulting solution is calculated using Raoult's law. In Part B, the total vapor pressure above the solution is calculated using the mole fractions and vapor pressures of each component.

Explanation:

Part A:

The vapor pressure of the resulting solution can be calculated using Raoult’s law, which states that the vapor pressure of a component in a solution is proportional to its mole fraction. The mole fraction of water can be calculated by dividing the moles of water by the total moles of solvent and solute. In this case, the mole fraction of water is 3.90 mol/(3.90 mol + 0.220 mol) = 0.946. Using Raoult’s law, the vapor pressure of the resulting solution is 0.946 times the vapor pressure of pure water, which is (0.946)(23.8 torr) = 22.5 torr.

Part B:

The total vapor pressure above the solution can be calculated using the mole fractions of each component and their respective vapor pressures. The mole fraction of cyclohexane is 1.00 mol/(1.00 mol + 2.40 mol) = 0.294, and the mole fraction of acetone is 2.40 mol/(1.00 mol + 2.40 mol) = 0.706. The total vapor pressure is the sum of the partial pressures of each component, which is (0.294)(97.6 torr) + (0.706)(229.5 torr) = 166.57 torr.

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