The solubility product of $AgCl$ is $4.0 \times 10^{-10}$ at $298 \ K$. The solubility of $AgCl$ in $0.04 \ m \ CaCl_2$ will be

  • A
    $2.0 \times 10^{-5} \ m$
  • B
    $1.0 \times 10^{-4} \ m$
  • C
    $5.0 \times 10^{-9} \ m$
  • D
    $2.2 \times 10^{-4} \ m$

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