The solubility of $AgCl$ in $0.1 \text{ M } NaCl$ is $S \text{ mol/L}$. If the solubility product of $AgCl$ is $1.8 \times 10^{-10}$, then $S$ is approximately:

  • A
    $1.8 \times 10^{-9} \text{ M}$
  • B
    $1.8 \times 10^{-10} \text{ M}$
  • C
    $1.8 \times 10^{-11} \text{ M}$
  • D
    $1.8 \times 10^{-12} \text{ M}$

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