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Question

In a 0.2 molar aqueous solution of a weak acid HX, the degree of ionization is 0.3. The freezing point of the solution will be nearest to

Change in freezing point = i * Kf * m

For dissociation of HX in water,

i = 1+degree of ionisation = 1+0.3=1.3

Kf for water = 1.86

m = 0.2 (given)

substitute value of i , Kf and m in the expression of change in freezing point, we get

change in F.P.= 1.3 * 1.86 * 0.2 = 0.4836 degree Celsius

hence new freezing point will be = 0 - 0.4836 = - 0.4836 degree Celsius

Thanks & Regards,

Nishant Srivastava

IIT JEE Faculty

(9860673153)

NISHANT SRIVASTAVA 8 year ago is this solution helpfull: 144 27