The equivalent weight fo `Na_2S_2O_3` in the reaction `2Na_2S_2O_3 + I_2 rarr Na_2 S_4O_6 + 2 NaI` will be .

Options

A. `M//2`
B. `M`
C. `M//0.5`
D. `M//8`

Explanation

To find the equivalent weight of \( \text{Na}_2\text{S}_2\text{O}_3 \) in the reaction

\[ 2\text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2\text{NaI} \]

we will follow these steps:

### Step 1: Determine the Molecular Weight of \( \text{Na}_2\text{S}_2\text{O}_3 \)

The molecular weight (M) of \( \text{Na}_2\text{S}_2\text{O}_3 \) can be calculated as follows:

- Sodium (Na): 23 g/mol, and there are 2 Na atoms, so \( 2 \times 23 = 46 \) g/mol
- Sulfur (S): 32 g/mol, and there are 2 S atoms, so \( 2 \times 32 = 64 \) g/mol
- Oxygen (O): 16 g/mol, and there are 3 O atoms, so \( 3 \times 16 = 48 \) g/mol

Adding these together:

\[ M = 46 + 64 + 48 = 158 \text{ g/mol} \]

### Step 2: Identify the Change in Oxidation State

In the reaction, we need to analyze the change in oxidation states of sulfur in \( \text{Na}_2\text{S}_2\text{O}_3 \) and \( \text{Na}_2\text{S}_4\text{O}_6 \).

1. In \( \text{Na}_2\text{S}_2\text{O}_3 \):
- The oxidation state of sulfur (S) can be calculated. Let the oxidation state of S be \( x \).
- The equation for \( \text{Na}_2\text{S}_2\text{O}_3 \) is:
\[ 2(23) + 2x + 3(-2) = 0 \implies 46 + 2x - 6 = 0 \implies 2x = -40 \implies x = -20 \implies \text{average oxidation state of S} = +2 \]

2. In \( \text{Na}_2\text{S}_4\text{O}_6 \):
- Similarly, let the oxidation state of S be \( y \).
- The equation for \( \text{Na}_2\text{S}_4\text{O}_6 \) is:
\[ 2(23) + 4y + 6(-2) = 0 \implies 46 + 4y - 12 = 0 \implies 4y = -34 \implies y = -8.5 \implies \text{average oxidation state of S} = +2.5 \]

### Step 3: Calculate the Change in Oxidation State

The change in oxidation state per sulfur atom from \( \text{Na}_2\text{S}_2\text{O}_3 \) to \( \text{Na}_2\text{S}_4\text{O}_6 \) is:

\[ \Delta \text{oxidation state} = 2.5 - 2 = 0.5 \]

Since there are 2 sulfur atoms in \( \text{Na}_2\text{S}_2\text{O}_3 \), the total change in oxidation state for 2 moles is:

\[ \text{Total change} = 2 \times 0.5 = 1 \]

### Step 4: Calculate the Equivalent Weight

The equivalent weight is given by the formula:

\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{Total change in oxidation state}} \]

Substituting the values:

\[ \text{Equivalent weight} = \frac{158 \text{ g/mol}}{1} = 158 \text{ g/equiv} \]

### Final Answer

The equivalent weight of \( \text{Na}_2\text{S}_2\text{O}_3 \) in the reaction is **158 g/equiv**.

---