In the section `:Na_(2)S_(2)O_(3)+4Cl_(2)+5H_(2)O rarr Na_(2)SO_(4)+H_(2)SO_(4)+8HCl`.
the equivalent weight of `Na_(2)S_(2)O_(3)` will be : `(M=` molecular weight of `Na_(2)S_(2)O_(3)`)

Options

A. `M//4`
B. `M//8`
C. `M//1`
D. `M//2`

Explanation

To find the equivalent weight of \( \text{Na}_2\text{S}_2\text{O}_3 \) in the given reaction, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction given is:
\[ \text{Na}_2\text{S}_2\text{O}_3 + 4\text{Cl}_2 + 5\text{H}_2\text{O} \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{SO}_4 + 8\text{HCl} \]

### Step 2: Determine the oxidation states
- For \( \text{Na}_2\text{S}_2\text{O}_3 \):
- Sodium (Na) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- Let the oxidation state of sulfur (S) be \( x \).

The oxidation state equation for \( \text{Na}_2\text{S}_2\text{O}_3 \) is:
\[ 2(+1) + 2x + 3(-2) = 0 \implies 2 + 2x - 6 = 0 \implies 2x = 4 \implies x = +2 \]

- For \( \text{Na}_2\text{SO}_4 \):
- The oxidation state equation is:
\[ 2(+1) + x + 4(-2) = 0 \implies 2 + x - 8 = 0 \implies x = +6 \]

### Step 3: Calculate the change in oxidation state
- The change in oxidation state of sulfur from \( +2 \) in \( \text{Na}_2\text{S}_2\text{O}_3 \) to \( +6 \) in \( \text{Na}_2\text{SO}_4 \) is:
\[ 6 - 2 = 4 \]
Since there are 2 sulfur atoms in \( \text{Na}_2\text{S}_2\text{O}_3 \), the total change in oxidation state is:
\[ 2 \times 4 = 8 \]

### Step 4: Calculate the equivalent weight
The equivalent weight (EW) of a substance in a redox reaction can be calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{NF}} \]
Where NF (number of electrons transferred) is 8 in this case.

Let \( M \) be the molecular weight of \( \text{Na}_2\text{S}_2\text{O}_3 \). Thus, the equivalent weight is:
\[ \text{Equivalent Weight} = \frac{M}{8} \]

### Final Answer
The equivalent weight of \( \text{Na}_2\text{S}_2\text{O}_3 \) is \( \frac{M}{8} \).

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