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The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase. |
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Answer» SOLUTION :Composition of Liquid mixture : Vapour pressure of pure liquid A `(p_(A)^(0))=450` mm Vapour presure of pure liquid B `(p_(B)^(0))=450 mm` Total vapour pressure of solution (P) = 600 mm According to Roult.s law : `p = p_(A)^(0)chi_(A)+p_(B)^(0)chi_(B)` `p=p_(A)^(0).chi_(A)+p_(B)^(0)(1-chi_(A))` `p=450.chi_(A)+700(1-chi_(A))` `600=450 chi_(A)+700-700 chi_(A)` `600-700=-250 chi_(A)` `chi_(A)=0.40` So, mole fraction of B = 0.60 Composition of vapour PHASE : `p_(A)=p_(A)^(0).chi_(A)=450xx0.40=180 mm` `p_(B)=p_(B)^(0).chi_(B)=700xx0.60=420 mm` Mole fraction of A in vapour `= (p_(A))/(p_(A)+p_(B))` `= (180)/(180+420)=0.30` So, mole fraction of B is = 0.70. |
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