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The vapour pressures 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. |
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Answer» Solution :Here, `p_(A)^(@)="450 mm,"p_(B)^(@)="700 mm,"p_("Total")="600 mm"` `"Applying Raoult's law,"p_(A)=x_(A)xxp_(A)^(@),""p_(B)=x_(B)xxp_(B)^(@)=(1-x_(A))p_(B)^(@)` `""P_("Total")=p_(A)+p_(B)=x_(A)p_(A)^(@)+(1-x_(A))p_(B)^(@)=p_(B)^(@)+(p_(A)^(@)-p_(B)^(@))x_(A)` Substituting the GIVEN values, we get `600=700+(450-700)x_(A)"or"250x_(A)=100"or"x_(A)=(100)/(250)=0.40` THUS, composition of the liquid mixture will be `x_(A)" (mole fraction of A) = 0.40,"x_(B)" (mole fractiobn of B)"=1-0.40=0.60` `THEREFORE""p_(A)=x_(A)xxp_(A)^(@)=0.40xx"450 mm = 180 mm,"p_(B)=x_(B)xxp_(B)^(@)=0.60xx"700 mm = 420 mm"` `"Mole fraction of A in the vapour phase"=(p_(A))/(p_(A)+p_(B))=(180)/(180+420)=0.30` Mole fraction of B in the vapour phaes `=1-0.30 = 0.70.` |
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