Correct answer: .. Solubility(s) = M 10 . 434 . 5 – 3    Ksp(AgCl) = (s)2 Ksp (AgCl) =10– 10 – log(Ksp) = 10

At 298 K, the | JEE Mains Chemistry Past Papers — Chem-Mantra

JEE Mains Chemistry Past PapershardNUMERICAL

Question

At 298 K, the

Answer: .

💡 Solution & Explanation

Solubility(s) = M 10 . 434 . 5 – 3    Ksp(AgCl) = (s)2 Ksp (AgCl) =10– 10 – log(Ksp) = 10

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