A3B2 is a sparingly — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

Solubility =       m x mole/lit A3B2 (s) 3A2+(aqs) + 2B3– (aqs) M x       M x Ksp = 3 M x M x             = (3)3 (2)2 m x       = 27 × 4 m x       = 108 m x       = M x a       So a = 108 Ans. | JEE MAIN-2021 | DATE : 31-08-2021 (SHIFT-1) | PAPER-1 | OFFICIAL BASED | CHEMISTRY PAGE # 11