In the reaction: `Na_(2)S_(2)O_(3)+4Cl_(2)+5H_(2)O rarr Na_(2)SO_(4)+H_(2)SO_(4)+8HCl`, the equivalent weight of `Na_(2)S_(2)O_(3)` will be: (M= molecular weight of `Na_(2)S_(2)O_(3))`

Options

A. (a)`M//4`
B. (b)`M//8`
C. (c )`M//1`
D. (d)`M//2`

Explanation

To find the equivalent weight of \( \text{Na}_2\text{S}_2\text{O}_3 \) in the given reaction, we need to follow these steps:

### Step 1: Identify the Reaction
The reaction provided is:
\[ \text{Na}_2\text{S}_2\text{O}_3 + 4\text{Cl}_2 + 5\text{H}_2\text{O} \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{SO}_4 + 8\text{HCl} \]

### Step 2: Determine the Change in Oxidation State
In this reaction, \( \text{Na}_2\text{S}_2\text{O}_3 \) is oxidized to \( \text{Na}_2\text{SO}_4 \) and \( \text{H}_2\text{SO}_4 \).

- The oxidation state of sulfur in \( \text{Na}_2\text{S}_2\text{O}_3 \) is +2.
- The oxidation state of sulfur in \( \text{Na}_2\text{SO}_4 \) and \( \text{H}_2\text{SO}_4 \) is +6.

### Step 3: Calculate the Change in Oxidation Number
Each sulfur atom changes from +2 to +6, which is an increase of 4. Since there are 2 sulfur atoms in \( \text{Na}_2\text{S}_2\text{O}_3 \), the total change in oxidation number is:
\[ \text{Total change} = 2 \times 4 = 8 \]

### Step 4: Determine the Number of Electrons Transferred
In redox reactions, the number of equivalents is equal to the total change in oxidation number. Therefore, the total number of electrons transferred is 8.

### Step 5: Calculate the Molecular Weight of \( \text{Na}_2\text{S}_2\text{O}_3 \)
To find the equivalent weight, we first need the molecular weight (M) of \( \text{Na}_2\text{S}_2\text{O}_3 \):
- Sodium (Na) = 23 g/mol, so \( 2 \times 23 = 46 \) g/mol
- Sulfur (S) = 32 g/mol, so \( 2 \times 32 = 64 \) g/mol
- Oxygen (O) = 16 g/mol, so \( 3 \times 16 = 48 \) g/mol

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

### Step 6: Calculate the Equivalent Weight
The equivalent weight is given by the formula:
\[ \text{Equivalent weight} = \frac{M}{n} \]
where \( n \) is the number of electrons transferred (which we found to be 8).

Thus,
\[ \text{Equivalent weight} = \frac{158}{8} = 19.75 \text{ g/equiv} \]

### 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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