Question

The total vapour pressure of an ideal solution obtained by mixing $3mol$ of 'A' and $7mol$ of 'B' at ${25}^{\circ }C$, is $297torr$.
(Given, Vapour pressure of pure 'A' at ${25}^{\circ }C$ is $150torr$)
Choose the correct option(s):

• A. The vapour pressure of pure B is $360torr$
• B. The mole fraction of A in vapour phase is 0.55
• C. The mole fraction of B in vapour phase is 0.52
• D. The mole fraction of B in vapour phase is 0.85

Answer 1

Answer: (A), (D)

The mole fraction of B in vapour phase is 0.85

$\text{Number of moles of A,}{n}_A=3mol\text{Number of moles of B,}{n}_B=7mol$

$\text{Mole fraction of A =}\frac{\text{No. of moles of A}}{\text{Total no. of moles of A and B}}$

$x_A=\frac{n_A}{n_A+n_B}=\frac{3}{3+7}=\frac{3}{10}=0.3x_B=1-0.3=0.7\text{By Raoult's law,}$
$p_A=x_A\times p_A^o$
$p_B=x_B\times p_B^o$
where,
$p_A^o$ is vapour pressure of pure A
$p_B^o$ is vapour pressure of pure B
$P_{total}=x_A\times p_A^o+x_B\times p_B^{o}297=150\times 0.3+p_B^{\circ }\times 0.7297-45=p_B^{\circ }\times 0.7$
$p_B^{\circ }=360torr$

Since,
$p_A=x_A\times p_A^o$
$y_A\times P_{total}=x_A\times p_A^o$
and
$y_B\times P_{total}=x_B\times p_B^o$
Here,
$y_A,y_B\text{are mole fractions of A and B in vapour phase}$
$y_A\times 297=0.3\times 150y_A=0.152y_B=1-y_A=1-0.152=0.848$
Options (a) and (d) are correct.