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Question

Benzene and toluene form ideal solution occur the entire range of composition The vapour pressure of pure benzene and toluene at 300K are 50:71 mm Hg and 32.06 mm Hg respectively calculate the mole fraction of benzene is vapour phase if 80g of benzene is mixed [ 100 g of toluene ].

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Solution

Number of moles of benzene= 80g78gmol1=1.026mole
Number of moles of toulene= 100g92gmol1=1.087mole
Mole fraction of benzene= 1.0261.026+1.087=0.486
Mole fraction of toulene= 10.486=0.514
Pbenzene=xbenzene×Pobenzene=0.486×50.71=24.65mm
Ptoulene=xtoulene×Potoulene=0.514×32.06=16.48mm
Mole fraction of benzene in vapour phase= PbenzenePbenzene+Ptoulene
=24.6524.65+16.48=0.60

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