1.

The vapour pressure of acetone at 20^(@)C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20^(@)C , its vapour pressure was 183 torr. The molar mass ( " g mol"^(-1)) of the substance is :

Answer»

32
64
128
488

Solution :Vap. pressure of PURE acetone, `p_(A)^(@)`
`=185` torr
Vap. pressure of solution, `p=183` torr
MOLAR mass of SOLVENT, `M_(A)=58 "g mol"^(-1)`,
`w_(B)=1.2 g, w_(A)=100 g`
`(p_(A)^(@)-p)/(p_(A)^(@))=x_(B)=(n_(B))/(n_(A))`
`(p_(A)^(@)-p)/(p_(A)^(@))=(w_(B))/(M_(B))xx(M_(A))/(w_(A))`
`(185-183)/(185)=(1.2)/(M_(B))xx(58)/(100)`
`(2)/(185)=(1.2xx58)/(M_(B)xx100)`
`:.M_(B)=(1.2xx58xx185)/(2xx100)=64.38`


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