At \(63.5^{\circ} \mathrm{C},\) the vapor pressure of \(\mathrm{H}_{2} \mathrm{O}\) is 175 torr, and that of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is 400 torr. A solution is made by mixing equal masses of \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) . (a) What is the mole fraction of ethanol in the solution? (b) Assuming ideal-solution behavior, what is the vapor pressure of the solution at \(63.5^{\circ} \mathrm{C} ?\) (c) What is the mole fraction of ethanol in the vapor above the solution?

Since the masses of water and ethanol in the solution are equal, we can denote them as \(m_{H_2O} = m_{C_2H_5OH} = m\). In order to calculate the mole fraction of ethanol in the solution, we need to first find the moles of both water and ethanol present in the solution. We can do this using the molecular weights of water and ethanol: Molecular weight of water: \(M_{H_2O} = 18.015 \,g/mol\) Molecular weight of ethanol: \(M_{C_2H_5OH} = 46.07 \,g/mol\) Moles of water: \(n_{H_2O} = \frac{m}{M_{H_2O}}\) Moles of ethanol: \(n_{C_2H_5OH} = \frac{m}{M_{C_2H_5OH}}\)

Now, we can calculate the mole fraction of ethanol, denoted as \(x_{C_2H_5OH}\), in the solution using the moles of water and ethanol: \(x_{C_2H_5OH} = \frac{n_{C_2H_5OH}}{n_{H_2O} + n_{C_2H_5OH}}\) Substituting the expressions for \(n_{H_2O}\) and \(n_{C_2H_5OH}\): \(x_{C_2H_5OH} = \frac{\frac{m}{M_{C_2H_5OH}}}{\frac{m}{M_{H_2O}} + \frac{m}{M_{C_2H_5OH}}}\) Since m appears in both numerator and denominator, we can cancel it out: \(x_{C_2H_5OH} = \frac{1/M_{C_2H_5OH}}{1/M_{H_2O} + 1/M_{C_2H_5OH}}\) Plug in the molecular weights: \(x_{C_2H_5OH} = \frac{1/46.07}{1/18.015 + 1/46.07} \approx 0.617\) The mole fraction of ethanol in the solution is approximately 0.617.

We will now calculate the vapor pressure of the solution using Raoult's law, which states that the partial pressure of a component in an ideal solution is equal to the mole fraction of the component multiplied by its vapor pressure: \(P_{solution} = x_{H_2O} P_{H_2O}^* + x_{C_2H_5OH} P_{C_2H_5OH}^*\) where: - \(P_{H_2O}^* = 175 \,torr\) and \(P_{C_2H_5OH}^* = 400 \,torr\) are the vapor pressures of water and ethanol at 63.5 °C - \(x_{H_2O} = 1 - x_{C_2H_5OH}\) is the mole fraction of water \(P_{solution} = (1 - 0.617)(175) + 0.617(400) = 0.383(175) + 0.617(400) \approx 295.12 \,torr\) The vapor pressure of the solution at 63.5 °C is approximately 295.12 torr.

Finally, we will calculate the mole fraction of ethanol in the vapor above the solution, denoted as \(y_{C_2H_5OH}\). This can be found using the partial pressures of water and ethanol in the vapor, which is equal to the product of mole fraction in the solution and the vapor pressure: \(P_{H_2O} = x_{H_2O} P_{H_2O}^* = 0.383(175) \approx 67.03 \,torr\) \(P_{C_2H_5OH} = x_{C_2H_5OH} P_{C_2H_5OH}^* = 0.617(400) \approx 246.8 \,torr\) Now, we can find the mole fraction of ethanol in the vapor: \(y_{C_2H_5OH} = \frac{P_{C_2H_5OH}}{P_{H_2O} + P_{C_2H_5OH}} = \frac{246.8}{67.03 + 246.8} \approx 0.786\) The mole fraction of ethanol in the vapor above the solution is approximately 0.786. In summary: (a) The mole fraction of ethanol in the solution is approximately 0.617. (b) The vapor pressure of the solution at 63.5 °C is approximately 295.12 torr. (c) The mole fraction of ethanol in the vapor above the solution is approximately 0.786.