Question

The total vapour pressure of an ideal solution obtained by mixing $4mol$ of 'A' and $6mol$ of 'B' at ${25}^{\circ }C$, is $210torr$. 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 $250torr$)

• A. 183.3
• B. 160
• C. 174.5
• D. 190.1

Answer 1

The correct option is A 183.3

$\text{Number of moles of A,}{n}_A=4mol\text{Number of moles of B,}{n}_B=6mol$

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

$x_A=\frac{n_A}{n_A+n_B}=\frac{4}{4+6}=\frac{4}{10}=0.4x_B=1-0.4=0.6\text{By Raoult's law,}$
$p_A=x_A\times p_A^o$
$p_B=x_B\times p_B^o$
where,
$p_A^o$ is vapour pressure of pure A
$p_B^o$ is vapour pressure of pure B​​​​​​​
$P_{total}=x_A\times p_A^o+x_B\times p_B^{o}210=250\times 0.4+p_B^{\circ }\times 0.6210=100+p_B^{\circ }\times 0.6$
$p_B^{\circ }=183.3torr$