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

The data on osmotic pressure is : `w_(2)=6 g " " V=250 cm^(3)=0.25` litre
`pi =4.5" atm, " T = 27+273=300 K`
R = 0.0821 litre atm/degree/mole
van't Hoff equation for osmotic for osmotic pressure is : `pi V = nRT`
`therefore 4.5xx0.25 = n xx0.0821xx300` or `n=(4.5xx0.25)/(0.0821xx300)=0.0457` mole
i.e., 0.457 mole of the solute are present in 250 mL of water (or 250 g of water).
`therefore` Molality of the solution `=(0.0457 mol)/(250 g)xx 1000 g kg^(-1)=0.1828" mol kg"^(-1)`, i.e., m = 0.1828
`K_(b)=0.52^(@)C//1000 "g (Given) " therefore Delta T_(b)=K_(b)xx m=0.52xx0.1828=0.095^(@)C`
`therefore` Boiling point of solution `(T_(b))=T_(b)^(@)+Delta T_(b)=100^(@)C+0.095^(@)=100.095^(@)C`