What volume of 0.500 M AgNO₃ is needed to precipitate all Cl⁻ in 25.0 mL of 0.100 M MgCl₂?
Q1. What volume of 0.500 M AgNO₃ is needed to precipitate all Cl⁻ in 25.0 mL of 0.100 M MgCl₂?
Answer: 10.0 mL
Explanation: Moles Cl⁻ = 0.025×0.100×2 = 0.005 → Volume = 0.005/(0.500) = 0.010 L. B uses 1:1 ratio.