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 step by step, we will 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 1: Determine the Normality of H₂SO₄
- The molarity (M) of H₂SO₄ is given as 0.1 M (decimolar).
- The n-factor for H₂SO₄ is 2 because it can donate 2 H⁺ ions.
- Normality (N) is calculated as:
\[ N_{H_2SO_4} = M \times n = 0.1 \times 2 = 0.2 \, N \]

### Step 2: Determine the Normality of NaOH
- The molarity (M) of NaOH is also given as 0.1 M (decimolar).
- The n-factor for NaOH is 1 because it can donate 1 OH⁻ ion.
- Normality (N) is calculated as:
\[ N_{NaOH} = M \times n = 0.1 \times 1 = 0.1 \, N \]

### Step 3: Calculate the Gram Equivalents of H₂SO₄ and NaOH
- The volume of both solutions is 50 ml, which we convert to liters for calculation:
\[ V_{H_2SO_4} = 50 \, ml = 0.050 \, L \]
\[ V_{NaOH} = 50 \, ml = 0.050 \, L \]
- Calculate the equivalents:
\[ \text{Equivalents of } H_2SO_4 = N_{H_2SO_4} \times V_{H_2SO_4} = 0.2 \times 0.050 = 0.01 \, \text{equivalents} \]
\[ \text{Equivalents of } NaOH = N_{NaOH} \times V_{NaOH} = 0.1 \times 0.050 = 0.005 \, \text{equivalents} \]

### Step 4: Determine the Excess Equivalents
- Since H₂SO₄ is a strong acid and NaOH is a strong base, they will react in a 1:1 ratio.
- The reaction will consume NaOH completely as it is the limiting reagent:
\[ \text{Excess equivalents of } H_2SO_4 = 0.01 - 0.005 = 0.005 \, \text{equivalents} \]

### Step 5: Calculate the Normality of the Resultant Solution
- The total volume of the resultant solution after mixing is:
\[ V_{total} = V_{H_2SO_4} + V_{NaOH} = 50 \, ml + 50 \, ml = 100 \, ml = 0.1 \, L \]
- The normality of the resultant solution is given by:
\[ N_{resultant} = \frac{\text{Excess equivalents}}{V_{total}} = \frac{0.005}{0.1} = 0.05 \, N \]

### Conclusion
The normality of the resultant solution is 0.05 N, confirming the assertion.