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At `80^(@)C`, the vapour p[ressure of pure liquid A is 520 mm of Hg and that of pure liquid B is 1000 mm of Hg. If a mixture solution of A and B boils at `80^(@)C` and 1 atomoshere pressure, the amount of A in the mixture is (1 atm = 760 mm of Hg)A. 60 mol precentB. 52 mol precentC. 34 mol presentD. 48 mol precent. |
Answer» Correct Answer - a According to available information, `P_(A)^(@)=520 mm Hg, P_(B)^(@)=1000mm Hg` `P_(A)^(@)X_(A)+P_(B)^(@)X_(B)=760 mm Hg`. `P_(A)^(@)X_(A)+P_(B)^(@)(1-X_(A))760` `520 X_(A)+1000(1-X_(A))=760` `520X_(A)+1000-1000X_(A)=760` `-480X_(A)=-240` or `X_(A)=240/480=1/2 "or 50 mol precent". |
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