Question

Calculate the vapor pressure of benzene, C6H6, at 25 °C in an ideal solution containing 1.76 mol of benzene and 0.435 mol of toluene. The vapor pressure of pure benzene is 96.0 torr.

Calculate the vapor pressure of toluene, C7H8, at 25 °C in an ideal solution containing 1.76 mol of benzene and 0.435 mol of toluene. The vapor pressure of pure toluene is 27.0 torr.

Answer 1

Answer: The FIRST> the vapor pressure of benzene in the solution at 25 °C is 77.1 torr.

Explanation: To solve this problem, we can use Raoult's law, which states that the vapor pressure of a component in an ideal solution is equal to the product of the mole fraction of that component and its vapor pressure in the pure state. The total vapor pressure of the solution is the sum of the partial pressures of the components. Mathematically, we can express this as:

P_total = P_A + P_B

where P_A and P_B are the vapor pressures of components A and B, respectively.

In this case, component A is benzene and component B is toluene. We are given the amount of each component in the solution, so we can calculate their mole fractions:

X_A = n_A / n_total = 1.76 mol / (1.76 mol + 0.435 mol) = 0.802

X_B = n_B / n_total = 0.435 mol / (1.76 mol + 0.435 mol) = 0.198

where n_total is the total amount of solute in the solution.

We are also given the vapor pressure of pure benzene, P_A°, which is 96.0 torr.

Using Raoult's law, we can calculate the vapor pressure of benzene in the solution:

P_A = X_A * P_A° = 0.802 * 96.0 torr = 77.1 torr

Finally, we can calculate the total vapor pressure of the solution:

P_total = P_A + P_B = 77.1 torr + 0 torr (since toluene is assumed to have zero vapor pressure) = 77.1 torr