# The solubility product of $AgCl$ is $4 \times 10^{-18}$ at $298\;k$ . The solublity of AgCl in $0.04\;MCaCl_2$ will be
$\begin{array}{1 1} (a)\;2 \times 10^{-4}\;M \\ (b)\;1 \times 10^{4}\;M \\ (c)\;5 \times 10^{-9}\;M \\ (d)\;2 \times 10^{-4}\;M\end{array}$