Conductivities of water and a saturated solution of a sparingly soluble salt AB₂ are 7 and 21 μ S m⁻¹, respectively. Given λA²⁺⁰=12.72 mS m² mol⁻¹ and λB²⁻⁰=7.64 mS m² mol⁻¹, the solubility of AB₂, in mol m⁻¹, is

Q62 · CSIR-NET Chemistry, December 2017

Paper: CSIR-NET December 2017 · Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Electrochemistry – General · Marks: 2 · Difficulty: Medium

Conductivities of water and a saturated solution of a sparingly soluble salt $\mathrm{AB}_{2}$ are 7 and $21 \mu \mathrm{S} \mathrm{m}^{-1}$, respectively. Given $\lambda_{\mathrm{A}^{2+}}^{0}=12.72 \mathrm{mS} \mathrm{m}^{2} \mathrm{mol}^{-1}$ and $\lambda_{\mathrm{B}^{2-}}^{0}=7.64 \mathrm{mS} \mathrm{m}^{2} \mathrm{mol}^{-1}$, the solubility of $\mathrm{AB}_{2}$, in mol $\mathrm{m}^{-1}$, is

(a)$5.0 \times 10^{-4}$

(b)$5.0 \times 10^{-3}$

(c)$5.0 \times 10^{-5}$

(d)$5.0 \times 10^{-6}$

Answer

Answer: A ✓ checked by 4AB · confidence high

Explanation

κ(salt) = 21 − 7 = 14 μS m⁻¹. Λ°(AB₂) = λ°(A²⁺) + 2λ°(B⁻) = 12.72 + 15.28 = 28.0 mS m² mol⁻¹. s = κ/Λ° = 14×10⁻⁶/28×10⁻³ = 5.0×10⁻⁴ mol m⁻³.

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