The solubility product of $AgCl$ is $1 \times 10^{-6}$ at $298 \, K$. Its solubility in $mol \, L^{-1}$ would be:
- A$1 \times 10^{-6} \, mol \, L^{-1}$
- B$1 \times 10^{-3} \, mol \, L^{-1}$
- C$1 \times 10^{-12} \, mol \, L^{-1}$
- DNone of these
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