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Question

A solute A dimerizes in water. The boiling point of a 2 molal solution of A is 100.52°C. The percentage association of A is (round off to the nearest integer) (use Kb for water =0.52 K kg mol-1; boiling point of water = 100°C) -


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Solution

Step 1: Elevation in boiling point for the solution -

Tb=Tb-Tb0 where, Tb is the boiling point of the solution and Tb0 is the boiling point of water.

Tb=100.52-100

Tb= 0.52

Step 2: Calculation of percentage association-

Tb= iKbm

Where, Kb(molal elevation constant) = 0.52 (given)

m is the molality of the solution= 2

Putting the values we get, 0.52 = i × 0.52 ×2

i = 0.520.52×2

i=1-α2

12=1-α2

α=1

Which means 100 % association.

So, the percentage association of A is 100 %.


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