1.

The vapour pressure of acetone at 20^(@)C is 185 torr. When 1.2 g of a non-volatile 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»

128
488
32
64

Solution :`(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1))=(w_(2)//M_(2))/(w_(1)//M_(1))=(185-183)/(185)=(1.2//M_(2))/(100//58)`
`("Molar mass of "CH_(3)COCH_(3)="58 G MOL"^(-1))`
`=(1.2xx58)/(100xxM_(2)) or M_(2)=(1.2xx58)/(100)xx(185)/(2)`
`="64.38 g mol"^(-1)`


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