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

A solution containing 6 g of a solute dissolved in 250 cm^(3) of water

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