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STANDARD Chemistry Colligative Properties MCQ Question

Type: MCQ-numerical-Medium-Class 12

The van't Hoff factor of 0.005 M aqueous solution of KCl is 1.95. The degree of ionisation of KCl is

A

0.95

B

0.97

C

0.94

D

0.96

Correct Answer

Option A

Detailed Explanation

KCl dissociates into K⁺ and Cl⁻ ions. The degree of ionisation (α) can be calculated using the formula α = (i - 1) / (n - 1), where i is the van't Hoff factor and n is the number of ions formed. Substituting the values, α = (1.95 - 1) / (2 - 1) = 0.95.

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