The solubility product of lead bromide is `8xx10^(-5)` at 298 K. If the salt is 80% dissociated in saturated solution, calculate the solubility of the salt (in gram/litre)

Explanation

To solve the problem, we need to calculate the solubility of lead bromide (PbBr2) in grams per liter, given that it is 80% dissociated in a saturated solution and its solubility product (Ksp) is \(8 \times 10^{-5}\) at 298 K.

### Step-by-Step Solution:

1. **Write the Dissociation Equation:**
The dissociation of lead bromide in water can be represented as:
\[
\text{PbBr}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{Br}^{-} (aq)
...