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 will follow these steps:

### Step 1: Determine the oxidation states of sulfur in hypo (Na₂S₂O₃)
In sodium thiosulfate (Na₂S₂O₃):
- Sodium (Na) has an oxidation state of +1.
- Let the oxidation state of sulfur be \( x \). Since there are two sulfur atoms, their total contribution is \( 2x \).
- Oxygen (O) has an oxidation state of -2, and there are three oxygen atoms, contributing -6.

Setting up the equation for the compound:
\[ 2(+1) + 2x + 3(-2) = 0 \]
\[ 2 + 2x - 6 = 0 \]
\[ 2x - 4 = 0 \implies 2x = 4 \implies x = +2 \]
Thus, the oxidation state of sulfur in Na₂S₂O₃ is +2.

### Step 2: Determine the oxidation states of sulfur in the products
**In Na₂SO₄:**
- Sodium (Na) = +1 (2 atoms contribute +2)
- Let the oxidation state of sulfur be \( x \).
- Oxygen (O) = -2 (4 atoms contribute -8).

Setting up the equation:
\[ 2(+1) + x + 4(-2) = 0 \]
\[ 2 + x - 8 = 0 \implies x - 6 = 0 \implies x = +6 \]
Thus, the oxidation state of sulfur in Na₂SO₄ is +6.

**In H₂SO₄:**
- Hydrogen (H) = +1 (2 atoms contribute +2)
- Let the oxidation state of sulfur be \( x \).
- Oxygen (O) = -2 (4 atoms contribute -8).

Setting up the equation:
\[ 2(+1) + x + 4(-2) = 0 \]
\[ 2 + x - 8 = 0 \implies x - 6 = 0 \implies x = +6 \]
Thus, the oxidation state of sulfur in H₂SO₄ is also +6.

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

### Step 4: Determine the number of sulfur atoms involved
In the reaction, there are 2 sulfur atoms in Na₂S₂O₃ that are oxidized, each changing from +2 to +6. Therefore, the total change in oxidation state for the two sulfur atoms is:
\[ \text{Total change} = 2 \times 4 = 8 \]

### Step 5: Calculate the equivalent weight
The equivalent weight of a substance is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{n-factor}} \]
Where the n-factor is the total change in oxidation state.

1. **Molecular weight of Na₂S₂O₃**:
- Na: 23 g/mol (2 Na = 46 g/mol)
- S: 32 g/mol (2 S = 64 g/mol)
- O: 16 g/mol (3 O = 48 g/mol)
- Total = 46 + 64 + 48 = 158 g/mol

2. **n-factor**: 8 (as calculated)

Now substituting these values:
\[ \text{Equivalent Weight} = \frac{158 \, \text{g/mol}}{8} = 19.75 \, \text{g/equiv} \]

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

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