The equivalent weight of `Na_(2)S_(2)O_(3)` ( Mol. wt = M ) in the reaction ,
`2Na_(2)S_(2)O_(3) + I_(2) rarr Na_(2) S_(4)O_(6) + 2NaI` is-

Options

A. `M//4`
B. `M//3`
C. `M//2`
D. Molecular mass

Answer

To find the equivalent weight of `Na₂S₂O₃` in the given reaction:

**Step 1: Write the balanced chemical equation.**
The balanced equation is:
\[ 2Na₂S₂O₃ + I₂ \rightarrow Na₂S₄O₆ + 2NaI \]

**Step 2: Determine the oxidation states of sulfur in the reactants and products.**
- In `Na₂S₂O₃`, let's denote the oxidation state of sulfur as \( x \).
- The equation for the oxidation state is:
\[ 2x + 2(+1) + 3(-2) = 0 \]
\[ 2x + 2 - 6 = 0 \]
\[ 2x - 4 = 0 \]
\[ 2x = 4 \]
\[ x = +2 \]

- In `Na₂S₄O₆`, we denote the oxidation state of sulfur as \( y \).
- The equation for the oxidation state is:
\[ 4y + 2(+1) + 6(-2) = 0 \]
\[ 4y + 2 - 12 = 0 \]
\[ 4y - 10 = 0 \]
\[ 4y = 10 \]
\[ y = +2.5 \]

**Step 3: Identify the change in oxidation states.**
- The oxidation state of sulfur changes from +2 in `Na₂S₂O₃` to +2.5 in `Na₂S₄O₆`.
- The change in oxidation state for sulfur is:
\[ \Delta = 2.5 - 2 = 0.5 \]

**Step 4: Determine the number of moles of electrons transferred (n-factor).**
- In the reaction, 2 moles of `Na₂S₂O₃` lose a total of:
\[ 2 \text{ moles} \times 0.5 \text{ (change in oxidation state)} = 1 \text{ mole of electrons} \]
- Therefore, the n-factor for `Na₂S₂O₃` is 1 (since 2 moles of `Na₂S₂O₃` correspond to 2 moles of sulfur, and each loses 0.5 electrons).

**Step 5: Calculate the equivalent weight.**
- The equivalent weight is given by the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{n \text{-factor}} \]
- Given that the molecular weight of `Na₂S₂O₃` is \( M \) and the n-factor is 1:
\[ \text{Equivalent weight} = \frac{M}{1} = M \]

Thus, the equivalent weight of `Na₂S₂O₃` in this reaction is \( M \).