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The solubility product of a sparingly soluble salt AX₂ is 3.2 ˣ 10⁻¹¹.Its solubility (in mol/L)is
Options
(a) 5.6 ˣ 10⁻⁶
(b) 3.1 ˣ 10⁻⁴
(c) 2 ˣ 10⁻⁴
(d) 4 ˣ 10⁻⁴
Correct Answer:
2 ˣ 10⁻⁴
Explanation:
K(sp) = 3.2×10⁻¹¹.
AX₂ ⇌ A² + 2X⁻
K(sp) = sx(2s)² = 4s³; i.e) 3.2×10⁻¹¹=4s³. (or) s³ = 0.8×10⁻¹¹
= 8×10⁻¹².
Therefore s = 2×10⁻⁴.
Related Questions: - In an isobaric process,when temperature changes from T₁ to T₂, ΔS is equal to
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Topics: Equilibrium
(104)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In an isobaric process,when temperature changes from T₁ to T₂, ΔS is equal to
- A corked flask containing boiling water and its vapour is allowed
- The oxidation state of sulpur in sodium tetrathionate (Na₂S₄O₆) is
- Which one of the elements has the highest ionisation energy?
- The structure of XeOF₄ is
Topics: Equilibrium (104)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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