

InterviewSolution
Saved Bookmarks
1. |
The solubility product of AgCl is `4.0xx10^(-10)` at 298 K. The solubility of AgCl in 0.04 m `CaCl_(2)` will beA. `2.0xx10^(-5) m`B. `1.0xx10^(-4)m`C. `5.0xx10^(-9)m`D. `2.2xx10^(-4)`m |
Answer» Correct Answer - C If x is the solubility of AgCl in 0.04 m `CaCl_(2)`, then `[Ag^(+)]= x "mol" L^(-1), [Cl^(-1)]=(0.04xx2)+x ~~0.8 m` `:. x (0.08 ) 4xx10^(-10) "or" x = 5.0 xx 10^(-9) m` |
|