1.

At 80^(@)C, the vapour pressure of pure liquid 'A' is 520 mm Hg and that of pure liquid 'B' is 1000 m Hg. If a mixtuce solution of 'A' and 'B' boils at 80^(@)C and 1 atm pressure, then amount of 'A' in the mixture is (1 atm = 760 mm Hg)

Answer»

48 MOL percent
50 mol percent
52 mol percent
34 mol percent

Solution :`P_("total")=(at 80^(@)C)=760mm`
`P_("total")=x_(A)p_(A)^(@)+x_(B)p_(B)^(@)=x_(A)p_(A)^(@)+(1-x_(A))p_(B)^(@)`
`=p_(B)^(@)+x_(A)(p_(A)^(@)-p_(B)^(@))`
`THEREFORE""1000+x_(A)(520-1000)=760`
`"or"480x_(A)=240`
`"or"x_(A)=0.50,` i.e., 50 mol percent.


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