A mixture of two volatile liquids A and B has a vapor pressure of 350 mm Hg at 300 K. The vapor pressure of B is 150 mm Hg. If the mole fraction of A is 0.4, its vapor pressure at same temperature will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 650 mm Hg
For a solution of two volatile liquids, Raoult’s law says the partial pressure of each component equals its mole fraction times the vapour pressure of the pure liquid. By Dalton’s law, the total vapour pressure is the sum of the partial pressures:
\( \displaystyle p_{\text{total}}=x_A\,p_A^{0}+x_B\,p_B^{0} \)
Here ptotal = 350 mm Hg, pB0 = 150 mm Hg, xA = 0.4 and so xB = 1 − 0.4 = 0.6:
\( \displaystyle 350=0.4\,p_A^{0}+0.6\times150 \)
\( \displaystyle 350=0.4\,p_A^{0}+90 \)
\( \displaystyle p_A^{0}=\frac{260}{0.4} \)
\( \displaystyle p_A^{0}=650\ \mathrm{mm\,Hg} \)
So the vapour pressure of pure A is 650 mm Hg.