In the neutralization of `Na_(2)S_(2)O_(3)` using `K_(2)Cr_(2)O_(7)` by idometry, the equivalent weight of `K_(2)Cr_(2)O_(7)` is

Options

A. `M//2`
B. `M//6`
C. `M//3`
D. `M`

Explanation

To find the equivalent weight of \( K_2Cr_2O_7 \) in the neutralization reaction with \( Na_2S_2O_3 \), we can follow these steps:

### Step 1: Understand the Reaction
The reaction between \( Na_2S_2O_3 \) (sodium thiosulfate) and \( K_2Cr_2O_7 \) (potassium dichromate) involves the reduction of chromium from its +6 oxidation state in \( K_2Cr_2O_7 \) to +3 oxidation state.

### Step 2: Write the Balanced Reaction
The balanced redox reaction can be represented as:
\[ 3 Na_2S_2O_3 + K_2Cr_2O_7 + 26 H^+ \rightarrow 6 SO_4^{2-} + 2 Cr^{3+} + 13 H_2O + 3 Na^+ \]

### Step 3: Determine the Change in Oxidation State
In \( K_2Cr_2O_7 \), chromium is in the +6 oxidation state and is reduced to +3 in the products. The change in oxidation state for each chromium atom is:
\[ +6 \text{ to } +3 = 3 \]
Since there are 2 chromium atoms in \( K_2Cr_2O_7 \), the total change in oxidation state is:
\[ 2 \times 3 = 6 \]

### Step 4: Calculate the n-factor
The n-factor is defined as the total number of moles of electrons transferred per mole of the compound in the reaction. Here, the n-factor for \( K_2Cr_2O_7 \) is 6.

### Step 5: Calculate the Molecular Weight of \( K_2Cr_2O_7 \)
The molecular weight (M) of \( K_2Cr_2O_7 \) can be calculated as follows:
- Potassium (K): 39.1 g/mol (2 K)
- Chromium (Cr): 52.0 g/mol (2 Cr)
- Oxygen (O): 16.0 g/mol (7 O)

Calculating the total:
\[ M = (2 \times 39.1) + (2 \times 52.0) + (7 \times 16.0) = 78.2 + 104.0 + 112.0 = 294.2 \text{ g/mol} \]

### Step 6: Calculate the Equivalent Weight
The equivalent weight (E) is calculated using the formula:
\[ E = \frac{M}{n} \]
Substituting the values we have:
\[ E = \frac{294.2 \text{ g/mol}}{6} = 49.03 \text{ g/equiv} \]

### Conclusion
The equivalent weight of \( K_2Cr_2O_7 \) is approximately **49.03 g/equiv**.

---