What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K. `K_(sp)` for `CaSO_(4)` is `9.0xx10^(-6)`.

Options

A. 2.45 L
B. 4.08 L
C. 4.90 L
D. 3.00 L

Explanation

To find the minimum volume of water required to dissolve 1 g of calcium sulfate (CaSO₄) at 298 K, given that the solubility product constant (Ksp) for CaSO₄ is \(9.0 \times 10^{-6}\), we can follow these steps:

### Step 1: Write the dissociation equation for calcium sulfate.
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).
Let the solubility of CaSO₄ in water be \(S\) mol/L. When 1 mole of CaSO₄ dissolves, it produces 1 mole of \(\text{Ca}^{2+}\) ions and 1 mole of \(\text{SO}_4^{2-}\) ions. Therefore, at equilibrium:
\[ [\text{Ca}^{2+}] = S \quad \text{and} \quad [\text{SO}_4^{2-}] = S \]

### Step 3: Write the expression for Ksp.
The solubility product constant \(K_{sp}\) is given by:
\[ K_{sp} = [\text{Ca}^{2+}][\text{SO}_4^{2-}] = S \cdot S = S^2 \]
Substituting the given \(K_{sp}\):
\[ S^2 = 9.0 \times 10^{-6} \]

### Step 4: Solve for S.
Taking the square root of both sides:
\[ S = \sqrt{9.0 \times 10^{-6}} = 3.0 \times 10^{-3} \text{ mol/L} \]

### Step 5: Calculate the molar mass of CaSO₄.
The molar mass of calcium sulfate (CaSO₄) is calculated as follows:
- Calcium (Ca): 40 g/mol
- Sulfur (S): 32 g/mol
- Oxygen (O): 16 g/mol × 4 = 64 g/mol

Adding these together:
\[ \text{Molar mass of CaSO}_4 = 40 + 32 + 64 = 136 \text{ g/mol} \]

### Step 6: Calculate the number of moles in 1 g of CaSO₄.
To find the number of moles in 1 g of CaSO₄:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \text{ g}}{136 \text{ g/mol}} \approx 0.00735 \text{ mol} \]

### Step 7: Relate moles to volume using solubility.
Using the solubility \(S\) we found earlier:
\[ S = 3.0 \times 10^{-3} \text{ mol/L} \]
To find the volume (V) required to dissolve 0.00735 moles of CaSO₄:
\[ \text{Volume} = \frac{\text{Number of moles}}{S} = \frac{0.00735 \text{ mol}}{3.0 \times 10^{-3} \text{ mol/L}} \approx 2.45 \text{ L} \]

### Conclusion
The minimum volume of water required to dissolve 1 g of calcium sulfate at 298 K is approximately **2.45 liters**.

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