The ionic strength of a solution containing 0.008 M AlCl₃ and 0.005 M KCl is

Q90 · CSIR-NET Chemistry, December 2016

Paper: CSIR-NET December 2016 · Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Debye Huckel · Marks: 2 · Difficulty: Easy

The ionic strength of a solution containing $0.008 \mathrm{M} \mathrm{AlCl}_{3}$ and 0.005 M KCl is

(a)0.134 M

(b)0.053 M

(c)0.106 M

(d)0.086 M

Answer

Answer: B ✓ checked by 4AB · confidence high

Explanation

I = ½[0.008×9 (Al³⁺) + 0.024×1 (Cl⁻) + 0.005×1 (K⁺) + 0.005×1 (Cl⁻)] = ½ × 0.106 = 0.053 M.

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