A solution of `Na_2S_2O_3` is standardised iodometrically by using `K_2Cr_2O_7`. The equivalent weight of `K_2Cr_2O_7` in this method is:

Options

A. Mol. Wt/2
B. Mol. Wt/6
C. Mol. Wt/3
D. equal to mol wt.

Explanation

To determine the equivalent weight of \( K_2Cr_2O_7 \) (potassium dichromate) when used iodometrically with \( Na_2S_2O_3 \) (sodium thiosulfate), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between potassium dichromate and sodium thiosulfate can be represented as follows:
\[ K_2Cr_2O_7 + 3 Na_2S_2O_3 + 6 H^+ \rightarrow 2 Cr^{3+} + 3 Na_2S_4O_6 + 7 H_2O \]
In this reaction, potassium dichromate is reduced from chromium in the +6 oxidation state to +3.

2. **Determine the Change in Oxidation State**:
The change in oxidation state for chromium in \( K_2Cr_2O_7 \) is from +6 to +3. The change in oxidation number per chromium atom is:
\[ 6 - 3 = 3 \]
Since there are 2 chromium atoms in \( K_2Cr_2O_7 \), the total change in oxidation state for the compound is:
\[ 2 \times 3 = 6 \]

3. **Calculate the n Factor**:
The n factor for \( K_2Cr_2O_7 \) is equal to the total change in oxidation state, which we found to be 6.

4. **Calculate the Molecular Weight of \( K_2Cr_2O_7 \)**:
The molecular weight of \( K_2Cr_2O_7 \) can be calculated as follows:
- Potassium (K): \( 39.1 \, \text{g/mol} \times 2 = 78.2 \, \text{g/mol} \)
- Chromium (Cr): \( 52.0 \, \text{g/mol} \times 2 = 104.0 \, \text{g/mol} \)
- Oxygen (O): \( 16.0 \, \text{g/mol} \times 7 = 112.0 \, \text{g/mol} \)
- Total molecular weight = \( 78.2 + 104.0 + 112.0 = 294.2 \, \text{g/mol} \)

5. **Calculate the Equivalent Weight**:
The equivalent weight of \( K_2Cr_2O_7 \) is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n} \]
Substituting the values:
\[ \text{Equivalent Weight} = \frac{294.2 \, \text{g/mol}}{6} = 49.03 \, \text{g/equiv} \]

6. **Select the Correct Option**:
The equivalent weight of \( K_2Cr_2O_7 \) can also be expressed as molecular weight divided by n factor. Since we found \( n = 6 \), the equivalent weight is:
- Option B: Molecular weight divided by 6.

### Final Answer:
The equivalent weight of \( K_2Cr_2O_7 \) in this method is **molecular weight divided by 6**.

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