Question
At $${80^ \circ }C,$$ the vapour pressure of pure liquid $$'A’$$ is $$520\,mm\,Hg$$ and that of pure liquid $$'B'$$ is $$1000\,mm\,Hg.$$ If a mixture solution of $$'A'$$ and $$'B'$$ boils at $${80^ \circ }C$$ and $$1\,atm$$ pressure, the amount of $$'A'$$ in the mixture is $$\left( {1\,atm = 760\,mm\,Hg} \right)$$
A.
$$52\,mol\,{\text{percent}}$$
B.
$$34\,mol\,{\text{percent}}$$
C.
$$48\,mol\,{\text{percent}}$$
D.
$$50\,mol\,{\text{percent}}$$
Answer :
$$50\,mol\,{\text{percent}}$$
Solution :
At 1 atmospheric pressure the boiling point of mixture is $${\text{8}}{{\text{0}}^ \circ }C.$$
At boiling point the vapour pressure of mixture, $${P_T} = 1$$ atmosphere $$ = 760\,mm\,Hg.$$
Using the relation,
$$\eqalign{
& {P_T} = P_A^ \circ {x_A} + P_B^ \circ {x_B},{\text{we get}} \cr
& {{\text{P}}_T} = 520{x_A} + 1000\left( {1 - {x_A}} \right) \cr
& \left\{ {\because \,P_A^ \circ = 520\,mm\,Hg,} \right. \cr
& P_B^ \circ = 1000\,mm\,Hg;\,{x_A} + {x_B} = 1 \cr
& {\text{or}}\,\,760 = 520{x_A} + 1000 - 1000{x_A}\,{\text{or}}\,\,{\text{480}}{{\text{x}}_A} = 240 \cr
& {\text{or}}\,\,{x_A} = \frac{{240}}{{480}} = \frac{1}{2}\,\,{\text{or}}\,\,50\,mol\,{\text{percent}} \cr} $$