JEE Main 2026 Chemistry question
Consider the following data.
| Electrolyte | Λ°m (S cm² mol⁻¹) | |---|---| | BaCl₂ | x₁ | | H₂SO₄ | x₂ | | HCl | x₃ |
BaSO₄ is sparingly soluble in water. If the conductivity of the saturated BaSO₄ solution is x S cm⁻¹ then the solubility product of BaSO₄ can be given as (Here Λm=Λ°m)
Answer: (A)
Solution
By Kohlrausch's law, Λ°m(BaSO₄)=Λ°(BaCl₂)+Λ°(H₂SO₄)−2Λ°(HCl)=x₁+x₂−2x₃. Solubility S=κ×1000/Λm=1000x/(x₁+x₂−2x₃) mol/L, so
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