The eq. wt. of `Na_(2)S_(2)O_(3)` as reductant in the reaction,
`Na_(2)S_(2)O_(3)+H_(2)O+Cl_(2)rarrNa_(2)SO_(4)+2HCl+S` is :

Options

A. (Mol. wt.)/1
B. (Mol. wt.)/2
C. (Mol. wt.)/6
D. (Mol. wt.)/8

Answer

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

### Step 1: Write the Balanced Reaction
The balanced reaction is:
\[ \text{Na}_2\text{S}_2\text{O}_3 + \text{H}_2\text{O} + \text{Cl}_2 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{HCl} + \text{S} \]

### Step 2: Determine the Oxidation States
1. **For \( \text{Na}_2\text{S}_2\text{O}_3 \)**:
- Let the oxidation state of sulfur (S) be \( x \).
- The equation for the oxidation state is:
\[ 2(+1) + 2x + 3(-2) = 0 \]
Simplifying gives:
\[ 2 + 2x - 6 = 0 \implies 2x - 4 = 0 \implies x = +2 \]
- So, the oxidation state of sulfur in \( \text{Na}_2\text{S}_2\text{O}_3 \) is \( +2 \).

2. **For \( \text{Na}_2\text{SO}_4 \)**:
- Let the oxidation state of sulfur be \( y \).
- The equation for the oxidation state is:
\[ 2(+1) + y + 4(-2) = 0 \]
Simplifying gives:
\[ 2 + y - 8 = 0 \implies y - 6 = 0 \implies y = +6 \]
- So, the oxidation state of sulfur in \( \text{Na}_2\text{SO}_4 \) is \( +6 \).

### Step 3: Calculate the Change in Oxidation State
The change in oxidation state for sulfur from \( +2 \) in \( \text{Na}_2\text{S}_2\text{O}_3 \) to \( +6 \) in \( \text{Na}_2\text{SO}_4 \) is:
\[ \Delta \text{Oxidation State} = +6 - (+2) = +4 \]

### Step 4: Determine the Number of Sulfur Atoms
In \( \text{Na}_2\text{S}_2\text{O}_3 \), there are 2 sulfur atoms.

### Step 5: Calculate the n-factor
The n-factor is calculated as:
\[ \text{n-factor} = \Delta \text{Oxidation State} \times \text{Number of S atoms} = 4 \times 2 = 8 \]

### Step 6: Calculate the Molecular Weight of \( \text{Na}_2\text{S}_2\text{O}_3 \)
The molecular weight of \( \text{Na}_2\text{S}_2\text{O}_3 \) is calculated as follows:
- Sodium (Na): \( 22.99 \, \text{g/mol} \times 2 = 45.98 \, \text{g/mol} \)
- Sulfur (S): \( 32.07 \, \text{g/mol} \times 2 = 64.14 \, \text{g/mol} \)
- Oxygen (O): \( 16.00 \, \text{g/mol} \times 3 = 48.00 \, \text{g/mol} \)

Adding these together:
\[ \text{Molecular Weight} = 45.98 + 64.14 + 48.00 = 158.12 \, \text{g/mol} \]

### Step 7: Calculate the Equivalent Weight
The equivalent weight is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{n-factor}} = \frac{158.12}{8} = 19.765 \, \text{g/equiv} \]

### Conclusion
Thus, the equivalent weight of \( \text{Na}_2\text{S}_2\text{O}_3 \) as a reductant in the reaction is approximately \( 19.77 \, \text{g/equiv} \).

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