Question

A mixture contains 1 mol of volatile liquid A (p$^\circ$A= 100 mmHg) and 3 moles of volatile liquid B (p$^\circ$b = 80 mmHg). If solution is ideally behaved, total vapour pressure of distillate is (approx)

• A. 85 mm Hg
• B. 86 mm Hg
• C. 90 mm Hg
• D. 92 mm Hg

Answer 1

Answer: (B)

$p_A = p^\circ _A \, x_A = 100 \times \frac{1}{4} $ = 25 mm Hg
$p_B = p^\circ _B \, x_B = 80 \times \frac{3}{4}$ = 60 mmHg
Mole fraction of A in vapour
$x'_A = \frac{p_A}{p_A + p_B} = \frac{25}{85}$
Mole fraction of B in vapour
$x'_B = 1 - \frac{25}{85}= \frac{60}{85}$
V.P. of distillate
= $x'_A p^\circ _A + x' _B p^\circ _B$
= $\frac{25}{85} \times 100 + \frac{60}{85} \times 80$
= $\frac{1}{85} ( 2500 + 4800)$
= $\frac{7300}{85} $ = 85.88 $\approx$ 86 mm Hg.