What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298. `K_(sp)` for `CaSO_4` is `9.1xx10^(-6)`.

Explanation

To find the minimum volume of water required to dissolve 1g of calcium sulfate (CaSO₄) at 298 K, we will follow these steps:

### Step 1: Write the dissociation equation for CaSO₄.
Calcium sulfate dissociates in water as follows:
\[ \text{CaSO}_4 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

### Step 2: Define the solubility (S) of CaSO₄.
Let the solubility of CaSO₄ be \( S \) moles per liter. Thus, at equilibrium:
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