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Solutions

Question
CBSEENCH12005603

The vapour pressure of pure liquids A and B are 450 and 700 mm Hg at 350 K respectively. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

Solution

It is given that:pA° = 450mm of HgpB0 = 700mm of Hgptotal = 600mm of Hgfrom Raoult's law , we have:pA = pA°xApB =pB0 xB  = pB°(1-xA)    therefore , total pressure, ptotal  = pA + pBptotal=   pA°xA + pB°(1-xA) ptotal =   pA°xA +  pB°-pB0 xA ptotal = ( pA° -  pB°)xA  +  pB°600= (450-750)xA + 700 -100 = -250xAxA = 0.4therefore ,xB  = 1-xA =1-0.4    =0.6Now, pA =   pA°xA               =450×0.4               =180 mm of HgpB  = pB0 xB      = 700×0.6      = 420mm of HgNow, fraction of liquid A  = pApA+pB          = 180180+420  = 180600  = 0.30

mole fraction of liquid b = 1-0.30 = 0.70