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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