JEE Advanced 2013 Chemistry question
The of is at 298 K. The solubility (in mol/L) of in a 0.1 M solution is
Answer: (B)
Solution
\mathrm{Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-}}, . In 0.1 M , Ag^+$}]
See the full step-by-step solution
Create a free account to read the complete working, and practise questions like this in a timed test.
Create a free account