In the following reaction (unbalanced), equivalent weight of `As_(2)S_(3)` is related to molecular weight `M` by
`As_(2)S_(3)+H+NO_(3)^(-) rarr NO+H_(2)O+AsO_(4)^(3-)+SO_(4)^(2-)`

Options

A. `(M)/(2)`
B. `(M)/(4)`
C. `(M)/(28)`
D. `(M)/(24)`

Explanation

To solve the problem, we need to determine the equivalent weight of \( As_2S_3 \) in the given reaction and how it relates to its molecular weight \( M \).

### Step-by-Step Solution:

1. **Identify the Reaction**:
The unbalanced reaction is:
\[ As_2S_3 + H^+ + NO_3^- \rightarrow NO + H_2O + AsO_4^{3-} + SO_4^{2-} \]

2. **Write the Products**:
From the reaction, we can see that arsenic (\( As \)) is being oxidized to arsenate (\( AsO_4^{3-} \)) and sulfur (\( S \)) is being oxidized to sulfate (\( SO_4^{2-} \)).

3. **Determine Oxidation States**:
- In \( As_2S_3 \):
- \( As \) has an oxidation state of +3.
- \( S \) has an oxidation state of -2.
- In the products:
- In \( AsO_4^{3-} \), \( As \) has an oxidation state of +5.
- In \( SO_4^{2-} \), \( S \) has an oxidation state of +6.

4. **Calculate Change in Oxidation States**:
- For \( As \):
- Change for 2 \( As \): \( (5 - 3) \times 2 = 4 \)
- For \( S \):
- Change for 3 \( S \): \( (6 - (-2)) \times 3 = (6 + 2) \times 3 = 8 \times 3 = 24 \)

5. **Total Change in Electrons**:
- Total electron transfer = Change from \( As \) + Change from \( S \):
\[ 4 + 24 = 28 \]

6. **Determine Valence Factor**:
- The valence factor is the total number of electrons transferred, which we calculated to be 28.

7. **Calculate Equivalent Weight**:
- The equivalent weight \( E \) is given by the formula:
\[ E = \frac{M}{\text{Valence Factor}} = \frac{M}{28} \]

### Final Answer:
The equivalent weight of \( As_2S_3 \) is related to its molecular weight \( M \) by:
\[ \text{Equivalent Weight} = \frac{M}{28} \]