A solution containing `6 g` of a solute dissolved in `250 cm^(3)` of water gave an osmotic pressure of `4.5 atm` at `27^(@)C`. Calculate the boiling point of the solution.The molal elevation constant for water is `0.52^@ C` per 1000 g.

Explanation

Van't Hoff factor for osmotic pressure is
`piV=nRT`
`4.5 xx 0.25=n xx 0.0821 xx300`
or `n=(4.5 xx 0.25)/(0.0821 xx300) =0.0457` mol
i.e., 0.0457 mol of the solute is present in `250 mL` of water or `250 g` of water.
`:.` Molality of the solution m =`(0.0457/250)xx1000`
`=0.1828`
`DeltaT_(b) = K_(b).m=0.52 xx 0.1828 = 0.095^(@)C`
`:.` Boiling point of solution
`(T_(s))=T_(0) + DeltaT_(b)=100 + 0.095 = 100.095^(@)C`