Question

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 masses : K = 39 u, Cl = 35.5 u). Assume,Potassium chloride is completely dissociated in solution.

Comments

Why we have added 100 at last step

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Here Tb is the difference in temperature so final boiling point is 100 degree+Tb

Answer 1

Molar mass of KCl = 39 + 35.5 = 74.35 g mol^-1 

As KCl dissociates completely, number of ions produced are 2.

Therefore, van't Hoff factor, i = 2 

Mass of KCl solution = 1000 x 1.04 = 1040 g

Mass of solvent = 1040 - 74.5 = 965.5 g

= 0.9655 kg

Molality of the solution :