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At 20^(@)C, The vapour pressure of liquid A is 22 mm of Hg and that of pure liquid B is 75 mm of Hg. What is thecomposition of these somponents in the solution that has the vapour presure of 48.5 mm of Hg at this tempreture ? (Assuming ideal solution behaviour). |
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Answer» `P_(A)=P_(A)^(@)+P_(A)=P_(A)^(@)x_(A)orP_(A)=(22mm)x_(A)` `P_(B)=P_(B)^(@)x_(B)orP_(B)=(22 mm)x_(B)=(75 mm)(1-x_(A))` According to availble data : `P_(A)+P_(B)=48.55 mm` `THEREFORE (22mm)x_(A)+(75 mm)(1-x_(A))=48.55 mm` `therefore (-53 mm)x_(A)=-26.45 mm` or `x_(A)((-26.45mm)/(-53mm))=0.5` Thus `x_(A)=0.5" and "x_(B)=(1-0.5)=0.5` `therefore "Two liquids have EQUAL moles of A and B".` |
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