The solubility product of ` PbCl_2 ` at 298 K is ` 7.1 xx 10 ^(-5) ` Calculate the solubility of ` PbCl_2 " in " g L^(-1) ` at 298 K.

Explanation

To calculate the solubility of lead chloride (PbCl₂) in grams per liter at 298 K, we can follow these steps:

### Step 1: Write the dissociation equation
Lead chloride dissociates in water as follows:
\[ \text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{Cl}^- (aq) \]

### Step 2: Define the solubility
Let the solubility of PbCl₂ be \( s \) moles per liter. This means:
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