Q39.The solubility of PbI2 at 25∘C is 0.7 g L−1 . The solubility product of PbI2 at this temperature is (molar mass of PbI2 = 461.2 g mol−1 ) (1) 1.40 × 10−9 (2) 0.14 × 10−9 (3) 140 × 10−9 (4) 14.0 × 10−9
What This Question Tests
This question requires converting solubility from mass per liter to molar solubility and then using the dissociation stoichiometry to calculate the solubility product constant (Ksp) for a sparingly soluble salt.
Concepts Tested
Formulas Used
s (mol/L) = s (g/L) / Molar mass
Ksp = [cation]^x [anion]^y
📚 NCERT Sections This Tests
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1.27 If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of CuS in aqueous solution.
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2.8 The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm–1. Calculate its molar conductivity.
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2.7 Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
📋 Question Details
- Chapter
- Ionic Equilibrium
- Topic
- Solubility product
- Year
- 2012
- Shift
- 26 May Online
- Q Number
- Q39
- Type
- MCQ
- NCERT Ref
- Class 11 Chemistry Ch 7: Equilibrium
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