Question

Liquids $A$ and $B$ form an ideal solution in the entire composition range. At $350K$, 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)

$P_A^{\circ }=7\times 10^3$
$P_B^{\circ }=12\times 10^3$
$x_A^{\prime }=0.4;x_B^{\prime }=1-0.4$
$x_B^{\prime }=0.6$
$P_{\text{total }}=P_A^{\circ }{x}_A^{\prime }+P_B^{\circ }{x}_B^{\prime }$
$=7\times 10^3\times 0.4+12\times 10^3\times 0.6$
$=(7\times 0.4+12\times 0.6)\times 10^3=10^4$
$x_A=\frac{P_A^{\circ }{x}_A^{\prime }}{P_{\text{total}}}=\frac{7\times 10^3\times 0.4}{10^4}$
$\therefore x_A=0.28,x_B=1-0.28=0.72$