Two formulal to calculate number of milli equivalents (mlQ)
Numbr of miliequivalents `=("weight")/("GEW")xx1000`
Numbr of milliequivalents= volume in ml `xx` Normality of solution
0.25 grams of pure `CaCO_(3)` neutralised 25 ml dilue HCl normality of HCl solution is

Options

A. 0.1 N
B. 0.5 N
C. 0.25 N
D. 0.2 N

Answer

To solve the problem of finding the normality of the HCl solution that neutralizes 0.25 grams of pure CaCO₃, we can follow these steps:

### Step 1: Calculate the molar mass of CaCO₃
The molar mass of calcium carbonate (CaCO₃) can be calculated by adding the atomic masses of its constituent elements:
- Calcium (Ca): 40 g/mol
- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol (there are 3 oxygen atoms)

\[ \text{Molar mass of CaCO₃} = 40 + 12 + (16 \times 3) = 40 + 12 + 48 = 100 \text{ g/mol} \]

### Step 2: Determine the n-factor for CaCO₃
The n-factor is determined by the number of positive ions multiplied by their valence. For CaCO₃:
- Calcium (Ca) has a valence of +2.
- The n-factor for CaCO₃ = 1 (Ca) × 2 = 2.

### Step 3: Calculate the equivalent mass of CaCO₃
The equivalent mass can be calculated using the formula:
\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{100}{2} = 50 \text{ g/equiv} \]

### Step 4: Calculate the number of milli equivalents of CaCO₃
Using the formula for milli equivalents:
\[ \text{Number of milli equivalents} = \frac{\text{Weight (g)}}{\text{Equivalent mass (g/equiv)}} \times 1000 \]
Substituting the values:
\[ \text{Number of milli equivalents of CaCO₃} = \frac{0.25}{50} \times 1000 = \frac{250}{50} = 5 \text{ mEq} \]

### Step 5: Set up the equation for HCl
Using the second formula for milli equivalents:
\[ \text{Number of milli equivalents} = \text{Volume (ml)} \times \text{Normality (N)} \]
Let the normality of HCl be \( n \). The volume of HCl is given as 25 ml:
\[ \text{Number of milli equivalents of HCl} = 25 \times n \]

### Step 6: Equate the milli equivalents for neutralization
For neutralization, the number of milli equivalents of CaCO₃ must equal the number of milli equivalents of HCl:
\[ 25n = 5 \]

### Step 7: Solve for normality (n)
\[ n = \frac{5}{25} = 0.2 \text{ N} \]

### Conclusion
The normality of the HCl solution is **0.2 N**.

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