NTA Abhyas JEE Main2020ChemistryIonic EquilibriumPractice
The Solubility of A B 2 is 0 .05 g per 100 mL at 25 o C . Calculate K sp of A B 2 at 25 o C? [Atomic mass of A = 20 amu, atomic mass of B = 40 amu ]
Options
- A1 0 - 3
- B5 × 1 0 - 7
- C1 0 - 6
- D5 × 1 0 - 3
Correct answer
B. 5 × 1 0 - 7
Step-by-step solution
= 0 .05 g/100 mL = 0 .5 g/L = 0.5 100 = 5 × 10 − 3   mol/L K s p = 4 s 3 = 4 × 5 × 1 0 - 3 3 = 5 × 1 0 - 7