Normality of 1 % `H_(2)SO_(4)` solution is nearly

Options

A. `2.5`
B. `0.1`
C. `0.2`
D. 1

Answer

To find the normality of a 1% \( H_2SO_4 \) solution, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the 1% Solution**:
A 1% \( H_2SO_4 \) solution means that there are 1 gram of \( H_2SO_4 \) in 100 grams of the solution.

2. **Finding the Volume of the Solution**:
Assuming the density of the solution is approximately 1 g/mL (which is a reasonable approximation for dilute solutions), the volume of 100 grams of solution is:
\[ \text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{100 \text{ g}}{1 \text{ g/mL}} = 100 \text{ mL} = 0.1 \text{ L} \]

3. **Calculating the Equivalent Weight of \( H_2SO_4 \)**:
The molecular weight of \( H_2SO_4 \) is approximately 98 g/mol. Since sulfuric acid can donate 2 protons (H\(^+\)), the equivalent weight is:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n}} = \frac{98 \text{ g/mol}}{2} = 49 \text{ g/equiv} \]

4. **Calculating the Number of Gram Equivalents**:
The number of gram equivalents of \( H_2SO_4 \) in 1 gram is:
\[ \text{Number of equivalents} = \frac{\text{mass}}{\text{equivalent weight}} = \frac{1 \text{ g}}{49 \text{ g/equiv}} \approx 0.0204 \text{ equiv} \]

5. **Calculating Normality**:
Normality (N) is defined as the number of equivalents per liter of solution:
\[ N = \frac{\text{Number of equivalents}}{\text{Volume in L}} = \frac{0.0204 \text{ equiv}}{0.1 \text{ L}} = 0.204 \text{ N} \]

6. **Final Approximation**:
To express this in a more manageable form, we can round it:
\[ N \approx 0.2 \text{ N} \]

### Conclusion:
The normality of a 1% \( H_2SO_4 \) solution is approximately **0.2 N**.