1.

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

Answer»

128
488
32
64

Solution :`(P_(A)^(@)P_(S))/P_(S)=(W_(B)xxW_(A))/(W_(A)xxM_(B))`
`((185-183))/((183))=((1.2 G)XX(58 g mol^(-1)))/((100g)xxM_(B))`
`M_(B)=((1.2g)xx(58g mol^(-1))xx(183))/(2xx(100g))=63.68g mol^(-1)`
=64 g `mol^(-1)`


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