Calculate the boiling point of a 1M aqueous solution (density 1.04 g `Ml^(-1)`) of Potassium chloride `(K_(b) " for water " =0.52 K kg "mol"^(-1)," Atomic masse " : K=39u, CI=39.9u)` Assume, Potassium chloride is completely dissociated in solution.

Explanation

Conc. Of the solution = 1 molar = 1 mol `L^(-1)`
Density of solution = 1.03 g `mL^(-1)`
Molar mass of NaCl = (23 + 35.5) g/mol = 58.5 g/mol So,
Mass of 1 litre of solution = 1000 `xx` 1.03 = 1030 g Therefore,
Mass of water containing 1 mole of NaCl = (1030 - 58.5)
g=971.5 g
Thus, Molality of the solution, m`=(1xx1000)/971.5mol//kg`
=1.0293 mol/kg
Then, `DeltaT_b=ixxk_bxxm=(2xx0.52xx1.0293)K`
1.07 K
So, Boiling point of solution (373.15 + 1.07) K
374.22 K