A: 50 ml, decinormal HCl when mixed with 50 ml, decinormal `H_(2) SO_(4)`, then normality of `H^(+)` ion in resultant solution is 0.1 N.
R: Here, `MV = M_(1)V_(1) - M_(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 \( H^+ \) ions in the resultant solution when 50 mL of decinormal \( HCl \) is mixed with 50 mL of decinormal \( H_2SO_4 \).

### Step-by-Step Solution:

1. **Understand the Normality of the Solutions**:
- Decinormal \( HCl \) means it has a normality of \( 0.1 \, N \) (1 decinormal = \( \frac{1}{10} \, N \)).
- Decinormal \( H_2SO_4 \) also has a normality of \( 0.1 \, N \).

2. **Calculate the Number of Equivalents**:
- For \( HCl \):
\[ \text{Normality} (N_1) = 0.1 \, N, \quad \text{Volume} (V_1) = 50 \, mL = 0.050 \, L \]
\[ \text{Equivalents of } HCl = N_1 \times V_1 = 0.1 \times 0.050 = 0.005 \, \text{equivalents} \]

- For \( H_2SO_4 \):
\[ \text{Normality} (N_2) = 0.1 \, N, \quad \text{Volume} (V_2) = 50 \, mL = 0.050 \, L \]
Since \( H_2SO_4 \) can donate 2 \( H^+ \) ions, the number of equivalents will be:
\[ \text{Equivalents of } H_2SO_4 = N_2 \times V_2 \times 2 = 0.1 \times 0.050 \times 2 = 0.01 \, \text{equivalents} \]

3. **Total Equivalents of \( H^+ \) Ions**:
- Total equivalents of \( H^+ \) ions from both acids:
\[ \text{Total } H^+ = 0.005 + 0.01 = 0.015 \, \text{equivalents} \]

4. **Calculate the Total Volume of the Solution**:
- The total volume after mixing:
\[ V_{total} = V_1 + V_2 = 50 \, mL + 50 \, mL = 100 \, mL = 0.1 \, L \]

5. **Determine the Normality of \( H^+ \) Ions in the Resultant Solution**:
- Normality (\( N \)) of \( H^+ \) ions:
\[ N = \frac{\text{Total equivalents}}{\text{Total volume in L}} = \frac{0.015}{0.1} = 0.15 \, N \]

6. **Final Result**:
- The normality of \( H^+ \) ions in the resultant solution is \( 0.15 \, N \).