What is the minimum volume of water required to dissolve `1.0 g` of calcium sulphate at `298 K`?
(For calcium sulphate , `K_(sp) is 9.1xx10^(-6))`.

Explanation

To find the minimum volume of water required to dissolve 1.0 g of calcium sulfate (CaSO₄) at 298 K, we will follow these steps:

### Step 1: Write the dissolution equation and Ksp expression
When calcium sulfate dissolves in water, it dissociates into its ions:
\[ \text{CaSO}_4 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

The solubility product (Ksp) expression for this equilibrium is:
\[ K_{sp} = [\text{Ca}^{2+}][\text{SO}_4^{2-}] \]

### Step 2: Define solubility (S)
Let the solubility of calcium sulfate be \( S \) mol/L. At equilibrium, the concentrations of the ions will be:
\[ [\text{Ca}^{2+}] = S \]
\[ [\text{SO}_4^{2-}] = S \]

### Step 3: Substitute into the Ksp expression
Substituting the concentrations into the Ksp expression gives:
\[ K_{sp} = S \times S = S^2 \]
Given that \( K_{sp} = 9.1 \times 10^{-6} \):
\[ S^2 = 9.1 \times 10^{-6} \]

### Step 4: Solve for solubility (S)
To find \( S \), take the square root of both sides:
\[ S = \sqrt{9.1 \times 10^{-6}} \]
\[ S \approx 3.02 \times 10^{-3} \, \text{mol/L} \]

### Step 5: Convert solubility to grams per liter
Next, we need to convert this solubility into grams per liter using the molar mass of calcium sulfate. The molar mass of CaSO₄ is approximately 136 g/mol.
\[ \text{Solubility in g/L} = S \times \text{Molar Mass} \]
\[ \text{Solubility in g/L} = 3.02 \times 10^{-3} \, \text{mol/L} \times 136 \, \text{g/mol} \]
\[ \text{Solubility in g/L} \approx 0.411 \, \text{g/L} \]

### Step 6: Calculate the volume of water needed for 1 g of CaSO₄
Now, we need to find the volume of water required to dissolve 1.0 g of CaSO₄. If 0.411 g of CaSO₄ requires 1 L of water, then:
\[ \text{Volume of water} = \frac{1.0 \, \text{g}}{0.411 \, \text{g/L}} \]
\[ \text{Volume of water} \approx 2.43 \, \text{L} \]

### Final Answer
The minimum volume of water required to dissolve 1.0 g of calcium sulfate at 298 K is approximately **2.43 liters**.

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