Let the solubility of AgCl in 0.1 M NaCl solution be $S$. The dissociation of AgCl is given by: $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$ In the solution, $\text{Cl}^-$ ions are also provided by the complete dissociation of the strong electrolyte NaCl. $[\text{Cl}^-] = S + 0.1 \approx 0.1 \text{ M}$ (since $S$ is very small compared to 0.1 M due to the common ion effect). $[\text{Ag}^+] = S$ The solubility product expression is: $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$ $1.6 \times 10^{-10} = S \times 0.1$ $S = \frac{1.6 \times 10^{-10}}{0.1} = 1.6 \times 10^{-9} \text{ M}$.