`K_("sp")` of `PbBr_(2)` (Molar mass `=367`) is `3.2xx10^(-5)`. If the salt is `80%` dissociated in solution, caculate the solubility of salt in `g` per `litre`.

Explanation

To solve the problem, we need to calculate the solubility of lead(II) bromide (PbBr₂) in grams per liter, given that it is 80% dissociated in solution and has a solubility product constant (Ksp) of 3.2 x 10^(-5) and a molar mass of 367 g/mol.

### Step-by-Step Solution:

1. **Write the dissociation equation for PbBr₂:**
\[ \text{PbBr}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2\text{Br}^- (aq) \]

2. **Define the solubility (S):**
Let the solubility of PbBr₂ be \( S \) mol/L. Since the salt is 80% dissociated, the concentrations of the ions at equilibrium will be:
- Pb²⁺: \( 0.80S \)
- Br⁻: \( 2 \times 0.80S = 1.60S \)

3. **Write the expression for Ksp:**
The solubility product (Ksp) is given by:
\[ K_{sp} = [\text{Pb}^{2+}][\text{Br}^-]^2 \]
Substituting the concentrations:
\[ K_{sp} = (0.80S)(1.60S)^2 \]

4. **Substitute Ksp value:**
Given \( K_{sp} = 3.2 \times 10^{-5} \):
\[ 3.2 \times 10^{-5} = (0.80S)(1.60S)^2 \]

5. **Simplify the equation:**
\[ 3.2 \times 10^{-5} = (0.80S)(2.56S^2) = 2.048S^3 \]

6. **Solve for S:**
\[ S^3 = \frac{3.2 \times 10^{-5}}{2.048} \]
\[ S^3 = 1.56 \times 10^{-5} \]
\[ S = (1.56 \times 10^{-5})^{1/3} \]
\[ S \approx 0.025 \text{ mol/L} \]

7. **Convert moles to grams:**
To find the solubility in grams per liter, use the molar mass:
\[ \text{Solubility (g/L)} = S \times \text{Molar Mass} \]
\[ \text{Solubility} = 0.025 \text{ mol/L} \times 367 \text{ g/mol} \]
\[ \text{Solubility} \approx 9.175 \text{ g/L} \]

### Final Answer:
The solubility of lead(II) bromide (PbBr₂) in grams per liter is approximately **9.175 g/L**.