Verified Solution • 2025–26 Session moderate

QJEEM08Ch60a9: JEE Main 2008 Chemistry - Class 12 JEEM08Ch60a9 Solution: At 80 o C, the vapour pressure of pure liquid ‘A’ is 520 mm Hg…

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)

Multiple Choice Options

  • A 52 mol percent
  • B 34 mol percent
  • C 48 mol percent
  • D 50 mol percent

Step-by-Step Verified Solution

At 11 atmospheric pressure the boiling point of mixture is 80∘C.{80^ \circ }C.

At boiling point the vapour pressure of mixture, PT=1{P_T} = 1

atmosphere =760 mm Hg.= 760\,mm\,Hg.

Using the relation,

PT=PA∘XA+PB∘XB,  {P_T} = P_A^ \circ {X_A} + P_B^ \circ {X_B},\,\, we get

PT=520XA+1000(1−XA){P_T} = 520{X_A} + 1000\left( {1 - {X_A}} \right)

{ \left\{ \, \right. PA∘=520mm  Hg,P_A^ \circ = 520mm\,\,Hg,

          \,\,\,\,\,\,\,\,\,\, PB∘=1000 mm Hg, XA+XB=1}\left. {P_B^ \circ = 1000\,mm\,Hg,\,{X_A} + {X_B} = 1} \right\}

or           \,\,\,\,\,\,\,\,\,\, 760=520XA+1000−1000XA  760 = 520{X_A} + 1000 - 1000{X_A}\,\,

or           \,\,\,\,\,\,\,\,\,\, 480XA=240480{X_A} = 240

or           \,\,\,\,\,\,\,\,\,\, XA=240480=12{X_A} = {{240} \over {480}} = {1 \over 2}

or           \,\,\,\,\,\,\,\,\,\, 5050 mol. percent

i.e., The correct answer is (d)(d)

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