Q43.The minimum volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution ( Ksp of PbCl2 = 3.2 × 10−8 ;atomic mass of Pb = 207 u ) is: (1) 1.798 L (2) 0.36 L (3) 17.98 L (4) 0.18 L
What This Question Tests
This question requires calculating the molar solubility from Ksp and then determining the minimum volume of water needed to dissolve a given mass of a sparingly soluble salt.
Concepts Tested
Formulas Used
Ksp = [Pb²⁺][Cl⁻]²
Solubility = mass / (Molar mass * Volume)
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📋 Question Details
- Chapter
- Ionic Equilibrium
- Topic
- Solubility product (Ksp)
- Year
- 2018
- Shift
- 15 Apr
- Q Number
- Q43
- Type
- Numerical
- NCERT Ref
- Class 11 Chemistry Ch 7: Equilibrium
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