Two liquids A and B form an ideal solution. At uss K, vapour pressure of solution containing 1 mol of A and 3 mol of B [K 550 mm of Hg. At the same temperature, if one more mol of B [K added to the mixture, the vapour pressure of the solution [K increased by 10 mm. The vapour pressure of A and B in the pure state, will be respectively.

First solution 

             $\begin{array}{l}\mathrm{P}=p_{\mathrm{A}}^0{x}_{\mathrm{A}}+p_{\mathrm{B}}^0{x}_{\mathrm{B}}\\ 550=\left(p_{\mathrm{A}}^0\times \frac{1}{1+3}\right)+\left(p_{\mathrm{B}}^0\times \frac{3}{1+3}\right)\\ \therefore x_{\mathrm{A}}=\frac{n_1}{n_1+n_2}\left.=\frac{1}{1+3}\right]\\ x_{\mathrm{B}}\left.=\frac{n_2}{n_1+n_2}=\frac{3}{1+3}\right]\end{array}$

 or       $550=p_{\mathrm{A}}^0\times 0.25+p_{\mathrm{B}}^0\times 0.75$       ….(1)

second solution  $P=p_A^0{x}_A+p_B^0{x}_B$

Where    $P=550+10=560\text{ torr }$

  $\begin{array}{c}560=\left(p_{\mathrm{A}}^0\times \frac{1}{4+1}\right)+p_{\mathrm{B}}^0\left(\frac{4}{4+1}\right)\\ \left(\because x_{\mathrm{A}}=\frac{n_1}{n_1+n_2}=\frac{1}{4+1};x_{\mathrm{B}}=\frac{n_2}{n_1+n_2}=\frac{4}{4+1}\right)\end{array}$

or $560=p_{\mathrm{A}}^0\times 0.2+p_{\mathrm{B}}^0\times 0.8 ....\left(2\right)$

Total number of mol in first solution 

 $=1+3=4$

 Total number of mol in second solution 

 $=1+(3+1)=5$

Multiplying equation (1) by 4 and (2) by 5 and subtracting ,

we get   

  $2200=p_{\mathrm{A}}^0\times 1+p_{\mathrm{B}}^0\times 3$

$\frac{2800=p_{\mathrm{A}}^0\times 1+p_{\mathrm{B}}^0\times 4}{-600\mathrm{mm}=-p_{\mathrm{B}}^0\text{ or }{p}_{\mathrm{B}}^0=600\mathrm{mm}.\text{ Ans. }}$

Substituting $p_{\mathrm{B}}^0=600\mathrm{mm}\text{ in (3), }$ we get 

             $2200=p_{\mathrm{A}}^0+3\times 600$

      $p_{\mathrm{A}}^0=2200-1800=400\mathrm{mm}$