The equlvalent wt. of `Na_(2)S_(2)O_(3)` in the reaction,
`Na_(2)S_(2)O_(3) + H_(2)O + Cl_(2) to Na_(2)SO_(4) + 2HCl + S ` , is

Options

A. `"mol. wt."/1`
B. `"mol. wt."/2`
C. `"mol. wt."/4`
D. `"mol. wt."/8`

Answer

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 is given as:
\[ \text{Na}_2\text{S}_2\text{O}_3 + \text{H}_2\text{O} + \text{Cl}_2 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{HCl} + \text{S} \]

### Step 2: Determine the oxidation states
We need to find the oxidation states of sulfur in \( \text{Na}_2\text{S}_2\text{O}_3 \) and \( \text{Na}_2\text{SO}_4 \).

1. **For \( \text{Na}_2\text{S}_2\text{O}_3 \)**:
- Let the oxidation state of sulfur be \( x \).
- The equation for the oxidation state is:
\[ 2x + 2(1) + 3(-2) = 0 \quad \text{(since Na is +1 and O is -2)} \]
- Simplifying gives:
\[ 2x + 2 - 6 = 0 \implies 2x - 4 = 0 \implies x = +2 \]

2. **For \( \text{Na}_2\text{SO}_4 \)**:
- The oxidation state of sulfur is \( +6 \) (since \( 2(+1) + x + 4(-2) = 0 \) leads to \( x = +6 \)).

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

### Step 4: Determine the number of moles of electrons transferred
- Since there are 2 sulfur atoms in \( \text{Na}_2\text{S}_2\text{O}_3 \), the total change in oxidation state for the two sulfur atoms is:
\[ n = \Delta \text{Oxidation State} \times \text{Number of sulfur atoms} = 4 \times 2 = 8 \]

### Step 5: Calculate the molecular weight of \( \text{Na}_2\text{S}_2\text{O}_3 \)
- The molecular weight (molar mass) of \( \text{Na}_2\text{S}_2\text{O}_3 \) can be calculated as follows:
- Sodium (Na): \( 23 \, \text{g/mol} \times 2 = 46 \, \text{g/mol} \)
- Sulfur (S): \( 32 \, \text{g/mol} \times 2 = 64 \, \text{g/mol} \)
- Oxygen (O): \( 16 \, \text{g/mol} \times 3 = 48 \, \text{g/mol} \)

Therefore, the total molecular weight is:
\[ \text{Molecular Weight} = 46 + 64 + 48 = 158 \, \text{g/mol} \]

### Step 6: Calculate the equivalent weight
- The equivalent weight of \( \text{Na}_2\text{S}_2\text{O}_3 \) is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n} = \frac{158 \, \text{g/mol}}{8} = 19.75 \, \text{g/equiv} \]

### Final Answer
The equivalent weight of \( \text{Na}_2\text{S}_2\text{O}_3 \) in the reaction is \( 19.75 \, \text{g/equiv} \).

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