Question

An ideal solution of 5.00 mole benzene and 3.25 mole toluene. At 298 K, the vapor pressure of the pure benzene and toluene were 96. Torr, and 28.9 Torr, respectively.
At what pressure does the vapor phase appear first, if the pressure above this solution is reduced from 760 Torr.

Answer 1

We can use Raoult's Law to determine the vapor pressure of each component in the solution. Raoult's Law states that the partial vapor pressure of a component in a solution is equal to the mole fraction of that component times its vapor pressure in its pure state.

First, let's find the mole fraction of each component in the solution:

Mole fraction of benzene (X_benzene) = moles of benzene / total moles in solution
X_benzene = 5.00 moles / (5.00 moles + 3.25 moles) = 5.00 / 8.25 ≈ 0.606

Mole fraction of toluene (X_toluene) = moles of toluene / total moles in solution
X_toluene = 3.25 moles / (5.00 moles + 3.25 moles) = 3.25 / 8.25 ≈ 0.394

Now we can apply Raoult's Law for each component:

Partial vapor pressure of benzene (P_benzene) = X_benzene * P_pure_benzene
P_benzene = 0.606 * 96.0 Torr ≈ 58.2 Torr

Partial vapor pressure of toluene (P_toluene) = X_toluene * P_pure_toluene
P_toluene = 0.394 * 28.9 Torr ≈ 11.4 Torr

The total vapor pressure of the solution (P_solution) is the sum of the partial vapor pressures of each component:

P_solution = P_benzene + P_toluene ≈ 58.2 Torr + 11.4 Torr ≈ 69.6 Torr

Therefore, the vapor phase will appear first when the pressure above the solution is reduced to approximately 69.6 Torr.