JEE MAIN - Chemistry (2008 - No. 25)
At 80oC, 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 80oC and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg)
52 mol percent
34 mol percent
48 mol percent
50 mol percent
Explanation
At $$1$$ atmospheric pressure the boiling point of mixture is $${80^ \circ }C.$$
At boiling point the vapour pressure of mixture, $${P_T} = 1$$
atmosphere $$ = 760\,mm\,Hg.$$
Using the relation,
$${P_T} = P_A^ \circ {X_A} + P_B^ \circ {X_B},\,\,$$ we get
$${P_T} = 520{X_A} + 1000\left( {1 - {X_A}} \right)$$
$$\left\{ \, \right.$$ $$P_A^ \circ = 520mm\,\,Hg,$$
$$\,\,\,\,\,\,\,\,\,\,$$ $$\left. {P_B^ \circ = 1000\,mm\,Hg,\,{X_A} + {X_B} = 1} \right\}$$
or $$\,\,\,\,\,\,\,\,\,\,$$ $$760 = 520{X_A} + 1000 - 1000{X_A}\,\,$$
or $$\,\,\,\,\,\,\,\,\,\,$$ $$480{X_A} = 240$$
or $$\,\,\,\,\,\,\,\,\,\,$$ $${X_A} = {{240} \over {480}} = {1 \over 2}$$
or $$\,\,\,\,\,\,\,\,\,\,$$ $$50$$ mol. percent
i.e., The correct answer is $$(d)$$
At boiling point the vapour pressure of mixture, $${P_T} = 1$$
atmosphere $$ = 760\,mm\,Hg.$$
Using the relation,
$${P_T} = P_A^ \circ {X_A} + P_B^ \circ {X_B},\,\,$$ we get
$${P_T} = 520{X_A} + 1000\left( {1 - {X_A}} \right)$$
$$\left\{ \, \right.$$ $$P_A^ \circ = 520mm\,\,Hg,$$
$$\,\,\,\,\,\,\,\,\,\,$$ $$\left. {P_B^ \circ = 1000\,mm\,Hg,\,{X_A} + {X_B} = 1} \right\}$$
or $$\,\,\,\,\,\,\,\,\,\,$$ $$760 = 520{X_A} + 1000 - 1000{X_A}\,\,$$
or $$\,\,\,\,\,\,\,\,\,\,$$ $$480{X_A} = 240$$
or $$\,\,\,\,\,\,\,\,\,\,$$ $${X_A} = {{240} \over {480}} = {1 \over 2}$$
or $$\,\,\,\,\,\,\,\,\,\,$$ $$50$$ mol. percent
i.e., The correct answer is $$(d)$$
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