JEE Main 2026 Chemistry question
The solubility product constants of Ag₂CrO₄ and AgBr are 32x and 4y respectively at 298 K. The value of can be expressed as:
Answer: (D)
Solution
For Ag₂CrO₄, = 4s₁³ = 32x ⇒ s₁ = . For AgBr, = s₂² = 4y ⇒ s₂ = 2√y. So s₁/s₂ = /√y.