The solubility of \(\mathrm{AgCl}\) is \(1 \times 10^{-5} \mathrm{~mol} / \mathrm{L}\). Its solubility in \(0.1\) molar sodium chloride solution is

\(K_{s p}\) of \(\mathrm{AgCl}=(\text { solubility of } \mathrm{AgCl})^{2}\)
\(
=\left(1 \times 10^{-5}\right)^{2}=1 \times 10^{-10}
\)
Suppose its solubility in \(0.1\) M \(\mathrm{NaCl}\) is \(x \mathrm{~mol} / \mathrm{L}\)
\(AgCl \rightleftharpoons Ag ^{+}+ Cl ^{-}\)
\(NaCl \rightleftharpoons Na ^{+}+\underset{0.1 M _{0.1 M }^{-}}{ Cl ^{-}}\)
\(\left[ Cl ^{-}\right]=(x+0.1) M\)
\(K_{s p}\) of \(AgCl =\left[ Ag ^{+}\right]\left[ Cl ^{-}\right]\)
\(=x \times(x+0.1)\)
\(1 \times 10^{-10}=x^2+0.1 x\)
Higher power of \(x\) are neglected
\(
\begin{aligned}
1 \times 10^{-10} &=0.1 x \\
x &=1 \times 10^{-9} \mathrm{M}
\end{aligned}
\)