Question

Liquids A and B form an ideal solution in the entire composition range. AT 350 K, the vapour 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, in equilibrium with a solution containing 40 mole percent of A at this temperature is:

• A. $y_A=0.28;y_B=0.72$
• B. $y_A=0.76;y_B=0.24$
• C. $y_A=0.37;y_B=0.63$
• D. $y_A=0.4;y_B=0.6$

Answer 1

The correct option is A $y_A=0.28;y_B=0.72$

Given,
Mole fraction of A in the solution = 0.4
Therefore, mole fraction of B in the solution = 0.6
Vapor pressure of pure A ($P_A^o$)= $7\times {10}^3$ Pa
Vapor pressure of pure B ($P_B^o$) = $12\times {10}^3$ Pa
$P_{total}=x_A{P}_A^{\circ }+x_B{P}_B^{\circ }$

$P_{total}=0.4\times 7\times {10}^3+0.6\times 12\times {10}^3$

Also mole fraction of A in the vapor phase is given by
$y_A\times P_{Total}=x_A\times P_A^0$
$\Rightarrow y_A=\frac{0.4\times 7\times {10}^3}{0.4\times 7\times {10}^3+0.6+12\times {10}^3}$
$\Rightarrow y_A=\frac{2.8}{10}=0.28$
$\therefore y_B=1-0.28=0.72$