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An aqueous solution (water is the solvent) of a KCl compound ( l = 2 ) boils at 102.5 degrees celcius. What is the molality (m) of the solution? Assume that the pure water boils at 100 degrees celcius.

Respuesta :

Explanation

Given:

i = 2

boiling point of water = 100 'C

KCl boiling point = 102.5 'C

Required: Molality (m) of the solution

Solution

DT = i x Kb x m

DT = 102.5-100 = 2.5 'C

DT = i x Kb x m

2.5 = 2 x 0.512 x m

2.5 = 1.024m

molality = 2.44 m

Answer

Molality = 2.44 m