At \(63.5^{\circ} \mathrm{C}\), the vapor pressure of $\mathrm{H}_{2} \mathrm{O}\( is \)23.3 \mathrm{kPa}\(, and that of ethanol \)\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\( is \)53.3 \mathrm{kPa}$. 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 idealsolution behavior, what is the vapor pressure of the solution at \(63.5^{\circ} \mathrm{C} ?(\mathbf{c})\) What is the mole fraction of ethanol in the vapor above the solution?

First, we are given equal masses of water and ethanol. We can find the moles of each substance by dividing the mass by the molar mass: For water (\(H_2O\)): Molar mass of \(H_2O = 18.02\,g/mol\) For ethanol (\(C_2H_5OH\)): Molar mass of \(C_2H_5OH = 46.07\,g/mol\) Let's assume we have 1 gram of each substance. Then, the moles of water and ethanol can be calculated as follows: Moles of \(H_2O = \frac{1}{18.02} = 0.0555\,mol\) Moles of \(C_2H_5OH =\frac{1}{46.07} = 0.0217\,mol\)

Now, we will calculate the mole fractions (\(x_i\)) of water and ethanol in the solution. The mole fraction of each component can be calculated using the formula: \(x_i = \frac{\text{moles of component i}}{\text{total moles in the solution}}\) Total moles in the solution = moles of \(H_2O\) + moles of \(C_2H_5OH = 0.0555 + 0.0217 = 0.0772\,mol\) Mole fraction of \(H_2O \,(x_{H_2O}) = \frac{0.0555\,\text{mol}}{0.0772\,\text{mol}} = 0.7185\) Mole fraction of \(C_2H_5OH \,(x_{C_2H_5OH}) = \frac{0.0217\,\text{mol}}{0.0772\,\text{mol}} = 0.2815\) (a) The mole fraction of ethanol in the solution is \(0.2815\).

As we are assuming ideal solution behavior, we can use Raoult's Law to calculate the vapor pressure of the solution. Raoult's Law states that the partial vapor pressure of a component in an ideal solution is equal to the product of the mole fraction of the component in the solution and its vapor pressure in the pure state: Partial vapor pressure of \(H_2O \,(P_{H_2O}) = x_{H_2O} \times P^{0}_{H_2O} = 0.7185 \times 23.3\,\text{kPa} = 16.73\,\text{kPa}\) Partial vapor pressure of \(C_2H_5OH \,(P_{C_2H_5OH}) = x_{C_2H_5OH} \times P^{0}_{C_2H_5OH} = 0.2815 \times 53.3\,\text{kPa} = 15.01\,\text{kPa}\) The total vapor pressure of the solution can be obtained by adding the partial vapor pressures: Total vapor pressure of the solution \(P = P_{H_2O} + P_{C_2H_5OH} = 16.73 + 15.01 = 31.74\,\text{kPa}\) (b) The vapor pressure of the solution at \(63.5^{\circ}C\) is \(31.74\,\text{kPa}\).

To find the mole fraction of ethanol in the vapor above the solution (\(y_{C_2H_5OH}\)), we can use the formula: \(y_{C_2H_5OH} = \frac{P_{C_2H_5OH}}{P} = \frac{15.01\,\text{kPa}}{31.74\,\text{kPa}} = 0.473\) (c) The mole fraction of ethanol in the vapor above the solution is \(0.473\).