Question
Two liquids $$A$$ and $$B$$ form ideal solutions. At $$300\,K,$$ the vapour pressure of a solution containing $$1\,mole$$ of $$A$$ and $$3\,mole$$ of $$B$$ is $$550\,\,mm\,\,Hg.$$ At the same temperature, if one more mole of $$B$$ is added to this solution, the vapour pressure of the solution increases by $$10\,\,mm\,\,Hg.$$ The vapour pressures of $$A$$ and $$B$$ in their pure states are respectively
A.
$$p_A^ \circ = 600\,\,mm\,\,Hg\,\,{\text{and}}\,\,p_B^ \circ = 400\,\,mm\,\,Hg$$
B.
$$p_A^ \circ = 550\,\,mm\,\,Hg\,\,{\text{and}}\,\,p_B^ \circ = 560\,\,mm\,\,Hg$$
C.
$$p_A^ \circ = 450\,\,mm\,\,Hg\,\,{\text{and}}\,\,p_B^ \circ = 650\,\,mm\,\,Hg$$
D.
$$p_A^ \circ = 400\,\,mm\,\,Hg\,\,{\text{and}}\,\,p_B^ \circ = 600\,\,mm\,Hg$$
Answer :
$$p_A^ \circ = 400\,\,mm\,\,Hg\,\,{\text{and}}\,\,p_B^ \circ = 600\,\,mm\,Hg$$
Solution :
Vapour pressure of solution containing $$1\,mole$$ of $$A + 3\,moles$$ of $$B = 550\,\,mm\,\,Hg$$
Vapour pressure of solution containing $$1\,mole$$ of $$A + 4\,moles$$ of $$B = \left( {550 + 10} \right) = 560\,\,mm\,\,Hg$$
$$\eqalign{
& {P_{{\text{Total}}}} = p_A^ \circ \times {x_A} + p_B^ \circ \times {x_B}\,\,{\text{or}}\,\,550 = p_A^ \circ \times {x_A} + p_B^ \circ \times {x_B} \cr
& = p_A^ \circ \times \frac{1}{4} + p_B^ \circ \times \frac{3}{4}\,\,\,\,\left[ {\because \,\,{x_A} = \frac{1}{{1 + 3}} = \frac{1}{4},{x_B} = \frac{3}{{1 + 3}} = \frac{3}{4}} \right] \cr
& 550 = \frac{{p_A^ \circ }}{4} + \frac{3}{4} \times p_B^ \circ \,\,\,{\text{or}}\,\,\,2200 = p_A^ \circ + 3p_B^ \circ \,\,\,\,\,\,\,\,\,\,\,...\left( {\text{i}} \right) \cr
& {\text{Again, we have}} \cr
& 560 = p_A^ \circ \times \frac{1}{5} + p_B^ \circ \times \frac{4}{5}\,\,\,\left( {\because \,\,{x_A} = \frac{1}{{1 + 4}} = \frac{1}{5},{x_B} = \frac{4}{{1 + 4}} = \frac{4}{5}} \right) \cr
& 2800 = p_A^ \circ + 4p_B^ \circ \,\,\,\,\,\,\,\,...\left( {{\text{ii}}} \right) \cr
& {\text{Solving equations (i) and (ii), we get}} \cr
& p_B^ \circ = 600\,\,mm\,\,Hg;\,\,p_A^ \circ = 400\,\,mm\,\,Hg \cr} $$