What is the volume of 0.1 N-HCl required to react completely with 1.0 gm of pure calcium carbonate

Options

A. `100cm^(3)`
B. `150cm^(3)`
C. `250cm^(3)`
D. `200cm^(3)`

Explanation

To find the volume of 0.1 N HCl required to react completely with 1.0 gm of pure calcium carbonate (CaCO₃), we can follow these steps:

### Step 1: Determine the Equivalent Weight of Calcium Carbonate
The equivalent weight of a compound is calculated as its molar mass divided by the number of equivalents of the reactive species. For calcium carbonate (CaCO₃):
- Molar mass of CaCO₃ = 40 (Ca) + 12 (C) + 3 × 16 (O) = 100 g/mol
- Since CaCO₃ reacts with HCl to produce CaCl₂, CO₂, and H₂O, it has one acidic hydrogen that reacts with HCl. Therefore, the equivalent weight is the same as the molar mass for this reaction.

**Equivalent weight of CaCO₃ = 100 g/mol**

### Step 2: Calculate the Number of Equivalents of CaCO₃
Using the formula for equivalents:
\[ \text{Number of equivalents} = \frac{\text{Weight of the substance (g)}}{\text{Equivalent weight (g/equiv)}} \]
Substituting the values:
\[ \text{Number of equivalents} = \frac{1.0 \text{ g}}{100 \text{ g/equiv}} = 0.01 \text{ equiv} \]

### Step 3: Use the Normality of HCl to Find the Volume Required
The relationship between normality (N), volume (V), and equivalents is given by:
\[ N \times V = \text{Number of equivalents} \]
Rearranging for volume (V):
\[ V = \frac{\text{Number of equivalents}}{N} \]
Substituting the values:
\[ V = \frac{0.01 \text{ equiv}}{0.1 \text{ N}} = 0.1 \text{ L} = 100 \text{ mL} \]

### Step 4: Convert Volume to cm³
Since 1 L = 1000 cm³, we convert 0.1 L to cm³:
\[ V = 0.1 \text{ L} \times 1000 \text{ cm³/L} = 100 \text{ cm³} \]

### Conclusion
The volume of 0.1 N HCl required to react completely with 1.0 gm of pure calcium carbonate is **100 cm³**.

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