The solubility (S) of a salt depends on its stoichiometry. For a salt $M_xA_y$, the solubility product is given by $K_{sp} = (xS)^x (yS)^y = x^x y^y S^{(x+y)}$.
For AgBr (1:1 type): $K_{sp} = S^2 \implies S = \sqrt{K_{sp}} = \sqrt{5.0 \times 10^{-13}} \approx 7.07 \times 10^{-7} \, M$.
For Zn(OH)2 (1:2 type): $K_{sp} = 4S^3 \implies S = \sqrt[3]{K_{sp}/4} = \sqrt[3]{1.0 \times 10^{-15} / 4} = \sqrt[3]{0.25 \times 10^{-15}} \approx 0.63 \times 10^{-5} = 6.3 \times 10^{-6} \, M$.
For Hg2Cl2 (2:2 or 1:1 simplified stoichiometry, but it dissociates as $Hg_2^{2+} + 2Cl^-$): $K_{sp} = 4S^3 \implies S = \sqrt[3]{1.3 \times 10^{-18} / 4} \approx \sqrt[3]{0.325 \times 10^{-18}} \approx 0.68 \times 10^{-6} = 6.8 \times 10^{-7} \, M$.
Comparing the calculated solubilities: $Zn(OH)_2 (6.3 \times 10^{-6}) > AgBr (7.07 \times 10^{-7}) > Hg_2Cl_2 (6.8 \times 10^{-7})$.
Final Answer: A
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