`1.0 g` of pure calcium carbonate was found to require `50 mL` of dilute `HCl` for complete reactions. The strength of the `HCl` solution is given by:

Options

A. `4N`
B. `2N`
C. `0.4N`
D. `0.2N`

Explanation

To find the strength of the HCl solution in terms of normality, we will follow these steps:

### Step 1: Determine the equivalent weight of calcium carbonate (CaCO₃).
- The molecular weight of calcium carbonate (CaCO₃) is calculated as follows:
- Calcium (Ca) = 40 g/mol
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol × 3 = 48 g/mol
- Therefore, the molar mass of CaCO₃ = 40 + 12 + 48 = 100 g/mol.

### Step 2: Determine the n-factor for calcium carbonate.
- Calcium carbonate (CaCO₃) reacts with hydrochloric acid (HCl) to produce calcium chloride (CaCl₂), water (H₂O), and carbon dioxide (CO₂).
- The reaction is:
\[ \text{CaCO}_3 (s) + 2 \text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{H}_2\text{O} (l) + \text{CO}_2 (g) \]
- From the reaction, we see that 1 mole of CaCO₃ reacts with 2 moles of HCl. Thus, the n-factor for CaCO₃ is 2.

### Step 3: Calculate the equivalent weight of CaCO₃.
- The equivalent weight is given by the formula:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{100 \text{ g/mol}}{2} = 50 \text{ g/equiv} \]

### Step 4: Calculate the number of gram equivalents of CaCO₃ in 1.0 g.
- The number of gram equivalents is calculated using the formula:
\[ \text{Number of equivalents} = \frac{\text{Given weight}}{\text{Equivalent weight}} = \frac{1.0 \text{ g}}{50 \text{ g/equiv}} = 0.02 \text{ equivalents} \]

### Step 5: Calculate the normality of the HCl solution.
- Normality (N) is defined as the number of equivalents of solute per liter of solution.
- We have 0.02 equivalents of CaCO₃ reacting in 50 mL of HCl solution. To find the normality:
\[ \text{Normality} = \frac{\text{Number of equivalents}}{\text{Volume of solution in liters}} = \frac{0.02 \text{ equivalents}}{0.050 \text{ L}} = 0.4 \text{ N} \]

### Conclusion:
The strength of the HCl solution is **0.4 N**.

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