Question

The minimum volume of water (in L) required to dissolve 1.5 g of \(CaSO_{4}\) (molar mass = 136 g mol\(^{-1}\)) at 298 K is given that \(K_{sp} = 9 \times 10^{-6}\).

• A. 6.37
• B. 7.37
• C. 3.67
• D. 3.73

Hint:

For AB-type salts: \(K_{sp} = s^2\). Always confirm dissociation type before solving.

Answer 1

The Correct Option is C

Concept: For a sparingly soluble salt \(CaSO_{4}\), dissociation is: \[ CaSO_{4}(s) \rightleftharpoons Ca^{2+} + SO_{4}^{2-} \] Thus, \[ K_{sp} = s^2 \]

Step 1: Calculate solubility \[ s = \sqrt{9 \times 10^{-6}} = 3 \times 10^{-3}\,\text{mol L}^{-1} \]

Step 2: Calculate moles of solute \[ n = \frac{1.5}{136} = 0.01103\,\text{mol} \]

Step 3: Calculate required volume \[ V = \frac{n}{s} = \frac{0.01103}{3 \times 10^{-3}} \approx 3.67\,\text{L} \]