100 mL of 0.1N `I_2` oxidizes `Na_2S_2O_3` in 50 ml solution to `Na_2S_4O_6`.The normality of this hypo solution would be :

Options

A. 0.1
B. 0.2
C. `1.0`
D. 1.6

Explanation

To solve the problem, we need to determine the normality of the sodium thiosulfate (Na₂S₂O₃) solution based on the reaction with iodine (I₂). We will use the concept of equivalents and the relationship between normality and volume.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Volume of I₂ solution (V₁) = 100 mL
- Normality of I₂ solution (N₁) = 0.1 N
- Volume of Na₂S₂O₃ solution (V₂) = 50 mL

2. **Calculate the Milliequivalents of I₂**:
- The formula for calculating milliequivalents is:
\[ \text{Milliequivalents} = \text{Normality} \times \text{Volume (in L)} \]
- Convert the volume of I₂ from mL to L:
\[ V₁ = 100 \, \text{mL} = 0.1 \, \text{L} \]
- Calculate the milliequivalents of I₂:
\[ \text{Milliequivalents of I₂} = N₁ \times V₁ = 0.1 \, \text{N} \times 0.1 \, \text{L} = 0.01 \, \text{equivalents} \]

3. **Use the Equivalence Principle**:
- According to the principle of equivalence, the milliequivalents of the oxidizing agent (I₂) will equal the milliequivalents of the reducing agent (Na₂S₂O₃):
\[ \text{Milliequivalents of I₂} = \text{Milliequivalents of Na₂S₂O₃} \]

4. **Set Up the Equation**:
- Let the normality of the Na₂S₂O₃ solution be \( N₂ \).
- The milliequivalents of Na₂S₂O₃ can be calculated as:
\[ \text{Milliequivalents of Na₂S₂O₃} = N₂ \times V₂ \]
- Substitute \( V₂ \) in liters:
\[ V₂ = 50 \, \text{mL} = 0.05 \, \text{L} \]
- Therefore, we have:
\[ 0.01 = N₂ \times 0.05 \]

5. **Solve for Normality (N₂)**:
- Rearranging the equation gives:
\[ N₂ = \frac{0.01}{0.05} = 0.2 \, \text{N} \]

### Final Answer:
The normality of the sodium thiosulfate (Na₂S₂O₃) solution is **0.2 N**.

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