A : 50 ml, decimolar `H_(2) SO_(4)` when mixed with 50 ml, decimolar NaOH, then normality of resultant solution is 0.05 N.
R: Here, `NV = |N_(1)V_(1) - N_(2) V_(2)|`

Options

A. If both Assertion & Reason are true and the reason is the correct explanation of the assertion
B. If both Assertion & Reason are true but the reason is not the correct explanation of the assertion
C. If Assertion is true statement but Reason is false
D. If both Assertion and Reason are false statements

Answer

To solve the problem, we need to determine the normality of the resultant solution when 50 mL of 0.1 M (decimolar) H₂SO₄ is mixed with 50 mL of 0.1 M NaOH.

### Step-by-Step Solution:

1. **Identify the Normality of H₂SO₄:**
- H₂SO₄ is a diprotic acid, meaning it can donate two protons (H⁺ ions).
- The normality (N) of H₂SO₄ can be calculated as:
\[ N_{H_2SO_4} = Molarity \times n \text{ (n = number of protons donated)} \]
\[ N_{H_2SO_4} = 0.1 \, \text{M} \times 2 = 0.2 \, \text{N} \]

2. **Identify the Normality of NaOH:**
- NaOH is a strong base that can donate one hydroxide ion (OH⁻).
- The normality of NaOH is:
\[ N_{NaOH} = Molarity \times n \]
\[ N_{NaOH} = 0.1 \, \text{M} \times 1 = 0.1 \, \text{N} \]

3. **Calculate the Total Volume of the Mixture:**
- The total volume after mixing is:
\[ V_{total} = V_{H_2SO_4} + V_{NaOH} = 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} \]

4. **Use the Formula for Resultant Normality:**
- The formula for the resultant normality (N_resultant) when mixing two solutions is:
\[ N_{resultant} = \frac{N_1V_1 - N_2V_2}{V_{total}} \]
- Here, \( N_1 = N_{H_2SO_4} = 0.2 \, \text{N} \), \( V_1 = 50 \, \text{mL} \), \( N_2 = N_{NaOH} = 0.1 \, \text{N} \), and \( V_2 = 50 \, \text{mL} \).

5. **Substituting the Values:**
- Substitute the values into the formula:
\[ N_{resultant} = \frac{(0.2 \, \text{N} \times 50 \, \text{mL}) - (0.1 \, \text{N} \times 50 \, \text{mL})}{100 \, \text{mL}} \]
- Calculate the numerator:
\[ N_{resultant} = \frac{(10 - 5)}{100} = \frac{5}{100} = 0.05 \, \text{N} \]

6. **Conclusion:**
- The normality of the resultant solution is 0.05 N.