Question

Liquids $A$ and $B$ form an ideal solution in the entire composition range. At $350\, K$, the vapor pressures of pure $A$ and pure $B$ are $7 \times 10^{3} Pa$ and $12 \times 10^{3} Pa$, respectively. The composition of the vapour is in equilibrium with a solution containing $40$ mole percent of $A$ at this temperature is:

• A. $x_{ A }=0.37 ; x_{ B }=0.63$
• B. $x_{ A }=0.28 ; x_{ B }=0.72$
• C. $x_{ A }=0.4 ; x_{ B }=0.6$
• D. $x_{ A }=0.76 ; x_{ B }=0.24$

Answer 1

Answer: (B)

$P _{ A }^{\circ}=7 \times 10^{3}$
$P _{ B }^{\circ}=12 \times 10^{3}$
$x_{ A }'=0.4 ; x_{ B }'=1-0.4$
$x_{ B }'=0.6$
$P _{\text {total }}= P^{\circ}_{A} x'_{A} + P^{\circ}_{B} x'_{B}$
$ = 7\times 10^3 \times 0.4 + 12 \times 10^3 \times 0.6$
$=( 7 \times 0.4 + 12 \times 0.6) \times 10^3 = 10^4$
$x_A = \frac{P^{\circ}_{A}x'_{A}}{P_{\text{total}}} = \frac{7\times 10^3 \times 0.4}{10^4}$
$\therefore x_A = 0.28, x_B = 1-0.28 = 0.72$