The solubility product of $AgCl$ at $25 \, ^\circ C$ is $5 \times 10^{-13}$. Its solubility is = .......
- A$5 \times 10^{-13}$
- B$7.1 \times 10^{-7}$
- C$2.5 \times 10^{-13}$
- D$2.5 \times 10^{-6}$
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