Two liquids \(A\) and \(B\) have vapor pressures of 76 \(\mathrm{mmHg}\) and \(132 \mathrm{mmHg}\), respectively, at \(25^{\circ} \mathrm{C}\). What is the total vapor pressure of the ideal solution made up of (a) 1.00 mole of \(A\) and 1.00 mole of \(B\) and (b) 2.00 moles of \(\mathrm{A}\) and 5.00 moles of \(\mathrm{B} ?\)

First, let's calculate the total moles present in the mixture for scenario (a): \[ total_moles = moles_A + moles_B = 1 + 1 = 2 \] The mole fraction of each component can be calculated as the moles of the component divided by total moles. So for A and B: \[ X_A = \frac{moles_A}{total_moles} = \frac{1}{2} = 0.5 \] \[ X_B = \frac{moles_B}{total_moles} = \frac{1}{2} = 0.5 \] According to Raoult's Law, the partial pressure of each component is the product of its mole fraction and its vapor pressure. So, for A and B: \[ P_A = X_A * P^0_A = 0.5 * 76 mmHg = 38 mmHg \] \[ P_B = X_B * P^0_B = 0.5 * 132 mmHg = 66 mmHg \] The total pressure is the sum of the partial pressures, so: \[ P_total = P_A + P_B = 38 mmHg + 66 mmHg = 104 mmHg \] So, the total vapor pressure of the solution in scenario (a) is 104 mmHg.

Now, let's calculate for scenario (b) following the same steps. First calculate total moles: \[ total_moles = moles_A + moles_B = 2 + 5 = 7 \] Then, calculate the mole fraction of each component: \[ X_A = \frac{moles_A}{total_moles} = \frac{2}{7} \approx 0.29 \] \[ X_B = \frac{moles_B}{total_moles} = \frac{5}{7} \approx 0.71 \] Next, find the partial pressure of each component using Raoult's Law: \[ P_A = X_A * P^0_A = 0.29 * 76 mmHg \approx 22 mmHg \] \[ P_B = X_B * P^0_B = 0.71 * 132 mmHg \approx 94 mmHg \] Finally, add these to find the total pressure: \[ P_total = P_A + P_B = 22 mmHg + 94 mmHg = 116 mmHg \] So, the total vapor pressure of the solution in scenario (b) is 116 mmHg.