Let the solubility of AgCl in 0.1 M NaCl solution be $S$. The dissociation of AgCl is given by: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$. Because NaCl is a strong electrolyte, it dissociates completely to give $[Cl^-] = 0.1 \text{ M}$. Due to the common ion effect, the additional $Cl^-$ from the dissolution of AgCl ($S$) is extremely small and can be neglected. Thus, total $[Cl^-] \approx 0.1 \text{ M}$. The solubility product expression is: $K_{sp} = [Ag^+][Cl^-]$ Substitute the given values into the equation: $1.6 \times 10^{-10} = S \times 0.1$ $S = \frac{1.6 \times 10^{-10}}{0.1} = 1.6 \times 10^{-9} \text{ M}$. Hence, the solubility of AgCl in 0.1 M NaCl is $1.6 \times 10^{-9} \text{ M}$.