Question

The total vapour pressure of an ideal solution obtained by mixing $1mol$ of 'A' and $2mol$ of 'B' at ${25}^{\circ }C$, is $180torr$. What is the vapour pressure (in $torr$) of pure 'B' at the same temperature?
(Given, Vapour pressure of pure 'A' at ${25}^{\circ }C$ is $150torr$)

• A. 150
• B. 125
• C. 55
• D. 195

Answer 1

The correct option is D 195

$\text{Number of moles of A,}{n}_A=1mol\text{Number of moles of B,}{n}_B=2mol$

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

$x_A=\frac{n_A}{n_A+n_B}=\frac{1}{3}{x}_B=1-\frac{1}{3}=\frac{2}{3}\text{By Raoult's law,}$
$p_A=x_A\times p_A^o$
$p_B=x_B\times p_B^o$
where,
$p_A^o$ is vapour pressure of pure A
$p_B^o$ is vapour pressure of pure B
$P_{total}=x_A\times p_A^o+x_B\times p_B^{o}180=\frac{1}{3}\times 150+\frac{2}{3}\times p_B^{\circ }$
$p_B^{\circ }=195torr$