The specific conductance at `25^(@)C` of a saturated solution of `SrSO_(4)` is `1.482 xx 10^(-4) ohm^(-1) cm^(-1)` while that of water used is `1.5 xx 10^(-6)` mho `cm^(-1)`. Determine at `25^(@)C` the solubiltiy in g per litre of `SrSO_(4)` in water. molar ionic conductance of `Sr^(2+)` and `SO_(4)^(2-)` ions at infinite are `54.46` and `79.8 ohm^(-1) cm^(2) "mole"^(-1)` respectively. `[Sr = 87.6 S = 32, O =16]`

Explanation

To determine the solubility of \( \text{SrSO}_4 \) in grams per liter, we will follow these steps:

### Step 1: Calculate the Specific Conductance of \( \text{SrSO}_4 \) Solution
The specific conductance of the saturated solution of \( \text{SrSO}_4 \) is given as:
\[ \kappa_{\text{solution}} = 1.482 \times 10^{-4} \, \text{ohm}^{-1} \text{cm}^{-1} \]
The specific conductance of water is given as:
\[ \kappa_{\text{water}} = 1.5 \times 10^{-6} \, \text{ohm}^{-1} \text{cm}^{-1} \]
To find the specific conductance of \( \text{SrSO}_4 \) alone, we subtract the conductance of water from that of the solution:
\[ \kappa_{\text{SrSO}_4} = \kappa_{\text{solution}} - \kappa_{\text{water}} = 1.482 \times 10^{-4} - 1.5 \times 10^{-6} = 1.467 \times 10^{-4} \, \text{ohm}^{-1} \text{cm}^{-1} \]

### Step 2: Calculate the Molar Ionic Conductance of \( \text{SrSO}_4 \)
Using Kohlrausch's law, the molar ionic conductance \( \lambda_m \) of \( \text{SrSO}_4 \) can be calculated as:
\[ \lambda_m = \lambda_{Sr^{2+}} + \lambda_{SO_4^{2-}} \]
Where:
- \( \lambda_{Sr^{2+}} = 54.46 \, \text{ohm}^{-1} \text{cm}^2 \text{mole}^{-1} \)
- \( \lambda_{SO_4^{2-}} = 79.8 \, \text{ohm}^{-1} \text{cm}^2 \text{mole}^{-1} \)

Calculating this gives:
\[ \lambda_m = 54.46 + 79.8 = 134.26 \, \text{ohm}^{-1} \text{cm}^2 \text{mole}^{-1} \]

### Step 3: Relate Conductance to Solubility
The relationship between the specific conductance \( \kappa \), molar ionic conductance \( \lambda_m \), and solubility \( S \) (in moles per liter) is given by:
\[ \kappa_{\text{SrSO}_4} = \lambda_m \cdot S \]
Rearranging this gives:
\[ S = \frac{\kappa_{\text{SrSO}_4}}{\lambda_m} \]
Substituting the values:
\[ S = \frac{1.467 \times 10^{-4} \, \text{ohm}^{-1} \text{cm}^{-1}}{134.26 \, \text{ohm}^{-1} \text{cm}^2 \text{mole}^{-1}} = 1.093 \times 10^{-6} \, \text{moles per liter} \]

### Step 4: Convert Moles to Grams
To convert the solubility from moles per liter to grams per liter, we need the molar mass of \( \text{SrSO}_4 \):
\[ \text{Molar mass of } \text{SrSO}_4 = 87.6 + 32 + (16 \times 4) = 183.6 \, \text{g/mol} \]
Now, converting moles to grams:
\[ \text{Solubility in g/L} = S \times \text{Molar mass} = 1.093 \times 10^{-6} \, \text{moles/L} \times 183.6 \, \text{g/mol} = 0.000201 \, \text{g/L} \]

### Final Answer
The solubility of \( \text{SrSO}_4 \) in water at \( 25^\circ C \) is approximately:
\[ \text{Solubility} \approx 0.201 \, \text{g/L} \]