Raoult's Law states that the partial vapour pressure of each component in an ideal solution is equal to the mole fraction of that component multiplied by vapour pressure of its pure component.
Given:
- Vapour pressure of A (\(P_A\)) = 21 kPa
- Vapour pressure of B (\(P_B\)) = 18 kPa
- Moles of A (\(n_A\)) = 1 mole
- Moles of B (\(n_B\)) = 2 moles
Total number of moles in the solution:
\(n_{\text{total}} = n_A + n_B = 1 + 2 = 3\)
Mole fractions of A and B:
\(\text{Mole fraction of A} (\chi_A) = \frac{n_A}{n_{\text{total}}} = \frac{1}{3}\)
\(\text{Mole fraction of B} (\chi_B) = \frac{n_B}{n_{\text{total}}} = \frac{2}{3}\)
Partial vapour pressures of A and B:
\(P_{\text{A solution}} = \chi_A \times P_A = \frac{1}{3} \times 21 = 7 \, \text{kPa}\)
\(P_{\text{B solution}} = \chi_B \times P_B = \frac{2}{3} \times 18 = 12 \, \text{kPa}\)
Finally, sum the partial vapor pressures to get the total vapor pressure of the mixture:
\(P_{\text{total}} = P_{\text{A solution}}\ +\ P_{\text{B solution}} = 7 + 12 = 19 \, \text{kPa}\)
Vapour pressure of the mixture is \(19 \, \text{kPa}\)
