Liquids A and B form an ideal solution at 30 °C. At this temperature, the vapour pressures of pure A and pure B are 200 mm Hg and 50 mm Hg respectively. Calculate the vapour pressure of the solution at the same temperature if equal numbers of moles of A and B are mixed. (Molar masses of A and B are 40 g mol−1 and 60 g mol−1 respectively.)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 125 mm Hg
For an ideal solution of two volatile liquids, Raoult’s law gives the partial pressure of each liquid as its mole fraction times its pure vapour pressure, and the total vapour pressure is their sum:
\( \displaystyle p_{\text{total}}=x_A\,p_A^{\circ}+x_B\,p_B^{\circ} \)
Equal numbers of moles of A and B are mixed, so the mole fractions are equal. The molar masses are not needed, because the amounts are already given in moles:
\( \displaystyle x_A=x_B=\frac{1}{2}=0.5 \)
\( \displaystyle p_{\text{total}}=0.5\times200+0.5\times50 \)
\( \displaystyle p_{\text{total}}=100+25=125\,\mathrm{mm\ Hg} \)