If the solubility product of lead iodide $(PbI_2)$ is $3.2 \times 10^{-8}$,then its solubility in $moles/litre$ will be

  • A
    $2 \times 10^{-3}$
  • B
    $4 \times 10^{-4}$
  • C
    $1.6 \times 10^{-5}$
  • D
    $1.8 \times 10^{-5}$

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