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Solutions

Question
CBSEENCH12005748

Heptane and Octane form an ideal solution at 373 K. The vapour pressures of the pure liquids at this temperature are 105.2 K Pa and 46.8 K Pa respectively. If the solution contains 25 g of heptane and 28.5 g of octane, calculate
(i) Vapour pressure exerted by heptane.
(ii) Vapour pressure exerted by solution.
(iii) Mole fraction of octane in the vapour phase.

Solution

Components A (Heptane, C7H16)
poA = 105.2 kpaWA = 25 gMA = (12×7) + (16×1) = 100
Components of B (octane, C8H18)
           poB = 46.8 kpaWA = 28.5 gMA = (12×8) + (18×1) = 114
No. of moles of heptane,
              nA = 25100=0.25
No. of moles of octane,
                nB =WBMB=28.5114=0.25
Total moles in solution,
               nA+nB = 0.25+0.25  = 0.50
(i) Vapour pressure exerted by heptane
                 pA = poA×A
where XA is mole fraction of component A
                   pA = 105.2×nAnA+nB = 105.2×0.250.50     = 52.6 k pa
(ii) Vapour pressure of octane
              pB = poB×BpB = 46.8 × nBnA+nB = 46.8×0.250.50       = 23.4 kpa
   Total vapour pressure of solution
                  p =pA+pB = 52.6+23.4 = 76 kpa
(iii) Mole fraction of octane
                              = nBnA+nB=0.250.50=0.5.