Question

Two moles of pure liquid 'A' $(P_A^0=80mm$ of Hg) and $3$ moles of pure liquid 'B' ($P_B^0=120mm$ of Hg) are mixed. Assuming ideal behaviour?

• A. Vapour pressure of the mixture is $104$mm of Hg
• B. Mole fraction of liquid 'A' in Vapour pressure is $0.3077$
• C. Mole fraction of 'B' in Vapour pressure is $0.692$
• D. Mole fraction of 'B' in Vapour pressure is $0.785$

Answer 1

The correct option is A Vapour pressure of the mixture is $104$mm of Hg

$P_A^{\circ }=80mmHg$ $P_B^{\circ }=120mmHg$

$n_A=2moles$ $n_B=3moles$

$x_A=\frac{2}{2+3}=\frac{2}{5}=0.4$

$x_B=\frac{3}{5}=0.6$

$P_A=P_A^{\circ }{x}_A=80\times 0.4=32mmHg$

$P_B=P_B^{\circ }{x}_B=120\times 0.6=72mmHg$

Total vapour pressure $P=P_A+P_B$

$=32+72=104mmHg$