The vapour pressure of pure liquid X and pure liquid Y at $250^\circ\text{C}$ are $120\text{ mm Hg}$ and $160\text{ mm Hg}$, respectively. If equal moles of X and Y are mixed to form an ideal solution, calculate the vapour pressure of the solution.

$140\text{ mm Hg}$

The correct answer is Option (2) → $140\text{ mm Hg}$ ##

$X = 120\text{ mm Hg}, P^\circ_{\text{A}} = P^\circ_{\text{X}} \cdot \chi_{\text{X}}$

$Y = 160\text{ mm Hg}, P^\circ_{\text{B}} = P^\circ_{\text{Y}} \cdot \chi_{\text{Y}}$

When equal amounts of X and Y are mixed

$\chi_{\text{A}} = \chi_{\text{B}} = 1/2 = 0.5$

According to Raoult's law, $P_{\text{Total}} = P_{\text{A}} + P_{\text{B}}$

$P^\circ_{\text{A}} \cdot \chi_{\text{A}} + P^\circ_{\text{B}} \cdot \chi_{\text{B}} = (1 - \chi_{\text{B}}) P^\circ_{\text{A}} + \chi_{\text{B}} P^\circ_{\text{B}}$

$P_{\text{Total}} = P^\circ_{\text{A}} + (P^\circ_{\text{B}} - P^\circ_{\text{A}}) \times \chi_{\text{B}}$

$= 120 + (160 - 120) \times 0.5$

$P_{\text{Total}} = 120 + 20 = 140\text{ mm Hg}$