In the estimation `Na_(2)S_(2)O_(3)`of using`Br_(2)` the equivalent weight of `Na_(2)S_(2)O_(3)` is :

Options

A. `M//2`
B. `M//8`
C. `M//3`
D. M

Explanation

To find the equivalent weight of \( \text{Na}_2\text{S}_2\text{O}_3 \) in the estimation using \( \text{Br}_2 \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between sodium thiosulfate (\( \text{Na}_2\text{S}_2\text{O}_3 \)) and bromine (\( \text{Br}_2 \)) in the presence of water can be represented as:
\[ \text{Na}_2\text{S}_2\text{O}_3 + \text{Br}_2 + \text{H}_2\text{O} \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{SO}_4 + \text{HBr} \]

### Step 2: Determine the oxidation states of sulfur
In \( \text{Na}_2\text{S}_2\text{O}_3 \):
- Let the oxidation state of sulfur be \( x \).
- The equation for the oxidation state is:
\[ 2(+1) + 2x + 3(-2) = 0 \]
This simplifies to:
\[ 2 + 2x - 6 = 0 \implies 2x = 4 \implies x = +2 \]
So, the oxidation state of sulfur in \( \text{Na}_2\text{S}_2\text{O}_3 \) is +2.

In \( \text{Na}_2\text{SO}_4 \):
- Let the oxidation state of sulfur be \( y \).
- The equation for the oxidation state is:
\[ 2(+1) + y + 4(-2) = 0 \]
This simplifies to:
\[ 2 + y - 8 = 0 \implies y = +6 \]
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 each sulfur atom is:
\[ \Delta \text{oxidation state} = +6 - (+2) = +4 \]
Since there are 2 sulfur atoms in \( \text{Na}_2\text{S}_2\text{O}_3 \), the total change is:
\[ \text{Total change} = 2 \times 4 = 8 \]
Thus, the valency factor (n factor) for \( \text{Na}_2\text{S}_2\text{O}_3 \) is 8.

### Step 4: Calculate the molecular weight of \( \text{Na}_2\text{S}_2\text{O}_3 \)
The molecular weight (M) of \( \text{Na}_2\text{S}_2\text{O}_3 \) can be calculated as follows:
- Sodium (Na): 23 g/mol, so for 2 Na: \( 2 \times 23 = 46 \) g/mol
- Sulfur (S): 32 g/mol, so for 2 S: \( 2 \times 32 = 64 \) g/mol
- Oxygen (O): 16 g/mol, so for 3 O: \( 3 \times 16 = 48 \) g/mol

Adding these together:
\[ M = 46 + 64 + 48 = 158 \text{ g/mol} \]

### Step 5: Calculate the equivalent weight
The equivalent weight (EW) 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
The equivalent weight of \( \text{Na}_2\text{S}_2\text{O}_3 \) is \( 19.75 \text{ g/equiv} \).

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