The solubility of $AgCl$ is $1.43 \times 10^{-3} \ g/L$. If the molar mass of $AgCl$ is $143 \ g/mol$,then the solubility product $(K_{sp})$ is:
- A$10^{-6}$
- B$10^{-8}$
- C$10^{-10}$
- D$10^{-5}$
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