Benzene and naphthalene form an ideal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at $$300$$ K are $$50.71$$ mm Hg and $$32.06$$ mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if $$80$$ g of benzene is mixed with $$100$$ g of naphthalene.

Question

Infinity Learn · Accepted Answer

nbenzene $$=80g78gmol$$−$$1=1.026mol$$   nnaphthalene $$=100128=0.781mol$$      xbenzene $$=1.0261.026+0.781=0.568xnaphthalene$$ $$=1$$−$$0.568=0.432$$  pbenzene $$=pbenzene$$ ∘×xberzene                     $$=50.71$$×$$0.568=28.80mmpnaphthalene$$ $$=32.06$$×$$0.432=13.85mmptotal$$ $$=28.80+13.85=42.65mmybenzene$$  in vapour phase $$=xbenzene$$ ×pbenzene ∘ptotal                                             $$=0.568$$×$$50.7142.65=0.675$$

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