Conductivity of saturated solution of `BaSO_(4)" at 315 K is "3.648 xx 10^(-6)" ohm"^(-1)" cm"^(-1)` and that of water is `1.25 xx10^(-6)" ohm"^(-1)" cm"^(-1)`. Ionic conductance of `Ba^(2+) and SO_(4)^(2-)" are 110 and 136.6 ohm"^(-1)" cm"^(2)" mol"^(-1)` respectively. Calculate the solubility of `BaSO_(4)` in g/L.

Explanation

To solve the problem of calculating the solubility of BaSO₄ in g/L, we will follow these steps:

### Step 1: Write down the given data
- Conductivity of saturated solution of BaSO₄ (κ) = \(3.648 \times 10^{-6} \, \text{ohm}^{-1} \text{cm}^{-1}\)
- Conductivity of water (κ_water) = \(1.25 \times 10^{-6} \, \text{ohm}^{-1} \text{cm}^{-1}\)
- Ionic conductance of Ba²⁺ (λ_Ba²⁺) = \(110 \, \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1}\)
- Ionic conductance of SO₄²⁻ (λ_SO₄²⁻) = \(136.6 \, \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1}\)

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