Q53.If the solubility product of PbS is 8 × 10−28 , then the solubility of PbS in pure water at 298 K is x × 10−16 mol L−1 . The value of x is____ (Nearest integer) [Given √2 = 1. 41]
What This Question Tests
This question tests the ability to calculate the solubility of a sparingly soluble salt from its given solubility product constant.
Concepts Tested
Formulas Used
Ksp = s²
📚 NCERT Sections This Tests
1.27 — If The Solubility Product Of Cus Is 6 × 10–16, Calculate The Maximum Molarity Of
Chemistry Class 11 · Chapter 1
1.27 If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of CuS in aqueous solution.
2.8 — The Conductivity Of 0.20 M Solution Of Kcl At 298 K Is 0.0248 S Cm–1. Calculate
Chemistry Class 11 · Chapter 2
2.8 The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm–1. Calculate its molar conductivity.
2.9 — The Resistance Of A Conductivity Cell Containing 0.001M Kcl Solution At 298
Chemistry Class 11 · Chapter 2
2.9 The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 W. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10–3 S cm–1. 59 Electrochemistry Reprint 2025-26
📋 Question Details
- Chapter
- Ionic Equilibrium
- Topic
- Solubility product
- Year
- 2022
- Shift
- 29 Jul Shift 1
- Q Number
- Q53
- Type
- Numerical
- NCERT Ref
- Class 11 Chemistry Ch 7: Equilibrium
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