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What is the minimum volume of water required to completely dissolve $10.0\text{ grams}$ of $Pb{Cl}_{2}(s)$?

$\left({K}_{sp, Pb{Cl}_{2}}=1.70\times {10}^{-5}\right)$

$8.73\text{ L of water}$

$8.99\times {10}^{-2}\text{ L of water}$

$1.03\text{ L of water}$

$2.22\text{ L of water}$