Calculate the boiling of one molar aqueous solution (density=1.04 g `mL^(-1)`) of potassium chloride `(K_(b)=0.52" K kg mol"^(-1))`

Explanation

Concentration of solution = 1M , Density of solution = 1.04 g `" mL"^(-1)`
Let us first calculate the motality of the solution .
mount of solute (KCI) = 1 mol = 74.5 g , Volume of solution = 1:L = 1000 mL
Mass of the solution ` = 1000 xx 1.04 g = 1040 g`
` :. ` Mass of solvent =` 1040 - 74.5` g = 965 . 5 g = 0.9655 kg
`"Molality of the solution "= ("No. of moles of the solute ")/("Mass of the solvent in kg")= ("1mol")/(0.9655 "kg") = 1.0357 "mol kg"^(-1) = 1.0357 m`
KCl dissociates as : ` KCl to K^(+)Cl^(-)`
` :. ` Number of particles after dissociation = 2 `:. ` van't Hoff factor , i = 2
Now ` Delta T _(b) = i xx K_(b) xx m = 2 xx 0.52 xx 1.0357 = 1.078^(@) C`
` :. ` Boiling point of the solution = ` 100 + 1.078 = 101.078 ^(@)C`