Question
The vapour pressure of two pure liquids $$A$$ and $$B$$ that form an ideal solution, are $$400$$ and $$800$$ $$mm$$ of $$Hg$$ respectively at a temperature $${t^ \circ }C.$$ The $$mole$$ fraction of $$A$$ in a solution of $$A$$ and $$B$$ whose boiling point is $${t^ \circ }C$$ will be
A.
0.4
B.
0.8
C.
0.1
D.
0.2
Answer :
0.1
Solution :
$$V.P.$$ of solution at $${t^ \circ }C = 760\,mm$$
[ at $$b.p.,$$ $$V.P.$$ of solution = atompheric pressure ]
$${\text{Thus}} = P_A^ \circ .\,{X_A} + P_B^ \circ .\,{X_B}$$
$${\text{or}}\,P = P_A^ \circ .\,{X_A} + P_B^ \circ .\left( {1 - {X_A}} \right)$$ $$\left[ {\because \,{X_A} + {X_B} = 1} \right]$$
$${\text{or}}\,\,760 = 400{X_A} + 800\left( {1 - {X_A}} \right)$$ $$\left[ {\because P = 760\,mm\,Hg} \right]$$
$$\eqalign{
& {\text{or}}\,\, - 800 + 760 = - 400\,{X_A} \cr
& {\text{or}}\,\, - 40 = - 400\,{X_A} \cr
& {\text{or}}\,\,{X_A} = \frac{{40}}{{400}} = 0.1 \cr} $$
Thus mole fraction of in solution is 0.1