One litre of a saturated solution of `CaCO_(3)` is evaporated to dryness due to which `7.0` mg of residue is left .
The solubility product for `CaCO_(3)` is

Options

A. `4.9 xx 10^(-8)`
B. `4.9 xx 10^(-5)`
C. `4.9 xx 10^(-9)`
D. `4.9 xx 10^(-7)`

Explanation

To find the solubility product (Ksp) of calcium carbonate (CaCO₃) from the given data, we can follow these steps:

### Step 1: Determine the mass of CaCO₃
The problem states that after evaporating 1 liter of a saturated solution of CaCO₃, 7.0 mg of residue is left.

### Step 2: Convert the mass of CaCO₃ to grams
Convert 7.0 mg to grams:
\[ 7.0 \, \text{mg} = 7.0 \times 10^{-3} \, \text{g} \]

### Step 3: Calculate the number of moles of CaCO₃
To find the number of moles, we need the molar mass of CaCO₃. The molar mass can be calculated as follows:
- Calcium (Ca) = 40 g/mol
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol × 3 = 48 g/mol

Thus, the molar mass of CaCO₃ is:
\[ \text{Molar mass of CaCO₃} = 40 + 12 + 48 = 100 \, \text{g/mol} \]

Now, we can calculate the number of moles:
\[ \text{Number of moles of CaCO₃} = \frac{\text{mass}}{\text{molar mass}} = \frac{7.0 \times 10^{-3} \, \text{g}}{100 \, \text{g/mol}} = 7.0 \times 10^{-5} \, \text{mol} \]

### Step 4: Determine the solubility (S) of CaCO₃
Since the solution volume is 1 liter, the solubility (S) in mol/L is equal to the number of moles:
\[ S = \frac{7.0 \times 10^{-5} \, \text{mol}}{1 \, \text{L}} = 7.0 \times 10^{-5} \, \text{mol/L} \]

### Step 5: Write the dissociation equation for CaCO₃
The dissociation of CaCO₃ in water can be represented as:
\[ \text{CaCO₃ (s)} \rightleftharpoons \text{Ca}^{2+} (aq) + \text{CO}_3^{2-} (aq) \]

### Step 6: Write the expression for the solubility product (Ksp)
The solubility product (Ksp) expression for CaCO₃ is given by:
\[ K_{sp} = [\text{Ca}^{2+}][\text{CO}_3^{2-}] \]
Since the solubility of CaCO₃ is S, we have:
\[ [\text{Ca}^{2+}] = S \quad \text{and} \quad [\text{CO}_3^{2-}] = S \]
Thus,
\[ K_{sp} = S \times S = S^2 \]

### Step 7: Calculate Ksp using the solubility value
Substituting the value of S:
\[ K_{sp} = (7.0 \times 10^{-5})^2 = 4.9 \times 10^{-9} \]

### Final Answer
The solubility product for CaCO₃ is:
\[ K_{sp} = 4.9 \times 10^{-9} \, \text{mol}^2/\text{L}^2 \]

---