The equiv. wt. of hypo in the reaction
`Na_(2) S_(2) O_(3) + Cl_(2) + H_(2) O rarr Na_(2) SO_(4) + H_(2) SO_(4) + HCl` is -

Options

A. `("Mol. wt")/(2)`
B. `("Mol. wt")/(4)`
C. `("Mol. wt")/(1)`
D. `("Mol. wt")/(8)`

Explanation

To find the equivalent weight of hypo (sodium thiosulfate, Na₂S₂O₃) in the given reaction, we can follow these steps:

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

### Step 2: Determine the Oxidation States
1. **For Sodium Thiosulfate (Na₂S₂O₃)**:
- Sodium (Na) has an oxidation state of +1.
- Let the oxidation state of sulfur (S) be \( x \).
- The oxidation state of oxygen (O) is -2.
- The equation for the oxidation states is:
\[ 2(+1) + 2x + 3(-2) = 0 \]
Simplifying gives:
\[ 2 + 2x - 6 = 0 \implies 2x - 4 = 0 \implies x = +2 \]
- Thus, the oxidation state of sulfur in Na₂S₂O₃ is +2.

2. **For Sodium Sulfate (Na₂SO₄)**:
- Sodium (Na) is +1, and the equation is:
\[ 2(+1) + x + 4(-2) = 0 \]
Simplifying gives:
\[ 2 + x - 8 = 0 \implies x = +6 \]
- The oxidation state of sulfur in Na₂SO₄ is +6.

3. **For Sulfuric Acid (H₂SO₄)**:
- The oxidation state of sulfur is also +6, as calculated similarly.

### Step 3: Calculate the Change in Oxidation State
- The change in oxidation state for sulfur from Na₂S₂O₃ to Na₂SO₄ (or H₂SO₄) is:
\[ \Delta \text{Oxidation State} = +6 - +2 = +4 \]

### Step 4: Determine the Number of Sulfur Atoms Involved
- There are 2 sulfur atoms in Na₂S₂O₃ that undergo oxidation.

### Step 5: Calculate the n-factor
- The n-factor is calculated as the total change in oxidation state multiplied by the number of atoms involved:
\[ n = \text{Change in Oxidation State} \times \text{Number of Atoms} = 4 \times 2 = 8 \]

### Step 6: Calculate the Molecular Weight of Hypo
- The molecular weight of Na₂S₂O₃ (sodium thiosulfate) can be calculated as follows:
- Na: 23 g/mol (2 atoms)
- S: 32 g/mol (2 atoms)
- O: 16 g/mol (3 atoms)
\[ \text{Molecular Weight} = (2 \times 23) + (2 \times 32) + (3 \times 16) = 46 + 64 + 48 = 158 \text{ g/mol} \]

### Step 7: Calculate the Equivalent Weight
- The equivalent weight is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{n-factor}} = \frac{158}{8} = 19.75 \text{ g/equiv} \]

### Final Answer
Thus, the equivalent weight of hypo (Na₂S₂O₃) in the reaction is **19.75 g/equiv**.

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