To determine which salt will precipitate last, we need to compare the solubility product constants ($K_{sp}$) of the salts.$\\ \\$Given $K_{sp}$ values are:$\\$• $K_{sp}($Ag$_2$CrO$_4) = 1.1 \times 10^{-12} \\$• $K_{sp}($AgCl$) = 1.8 \times 10^{-10} \\$• $K_{sp}($AgBr$) = 5.0 \times 10^{-13} \\$• $K_{sp}($AgI$) = 8.3 \times 10^{-17} \\ \\$The salt with the highest $K_{sp}$ will precipitate last because it is the most soluble.$\\ \\$Let's compare the $K_{sp}$ values:$\\$• $K_{sp}($AgCl$) = 1.8 \times 10^{-10} \\$• $K_{sp}($AgBr$) = 5.0 \times 10^{-13} \\$• $K_{sp}($Ag$_2$CrO$_4) = 1.1 \times 10^{-12} \\$• $K_{sp}($AgI$) = 8.3 \times 10^{-17} \\ \\$Among these, $K_{sp}($Ag$_2$CrO$_4) = 1.1 \times 10^{-12}$ is the highest after $K_{sp}($AgCl).$\\$However, since Ag$_2$CrO$_4$ has a different stoichiometry, we need to consider its solubility.$\\ \\$For $$Ag$_2$CrO$_4,$ the dissolution is:$\\$Ag$_2$CrO$_4 \rightleftharpoons 2$Ag$^+ +$CrO$_4^{2-} \\$Let the solubility of $$Ag$_2$CrO$_4$ be $s. \\$Then, $[$Ag$^+] = 2s$ and $[$CrO$_4^{2-}] = s. \\ \\ K_{sp} = (2s)^2 \cdot s = 4s^3 \\ 1.1 \times 10^{-12} = 4s^3 \\ s^3 = \frac{1.1 \times 10^{-12}}{4} \\ s^3 = 2.75 \times 10^{-13} \\ s = (2.75 \times 10^{-13})^{1/3} \\ s \approx 6.5 \times 10^{-5} \\ \\$For $$AgCl, the dissolution is:$\\$AgCl$\rightleftharpoons$Ag$^+ +$Cl$^- \\ K_{sp} = s^2 \\ s = \sqrt{1.8 \times 10^{-10}} \\ s \approx 1.34 \times 10^{-5} \\ \\$Comparing solubilities, $$Ag$_2$CrO$_4$ has a higher solubility than $$AgCl.$\\ \\$Therefore, $$Ag$_2$CrO$_4$ will precipitate last.$\\$This corresponds to Option 3.