In the standardization of `Na_(2)S_(2)O_(3)` using `K_(2)Cr_(2)O_(7)` by iodometry, the equivalent weight of `K_(2)Cr_(2)O_(7)` is `("molecular weight")/(X)`. What is the vlaue of X.

Explanation

To find the value of \( X \) in the expression for the equivalent weight of \( K_2Cr_2O_7 \) in the iodometric standardization with \( Na_2S_2O_3 \), we need to determine the n-factor for \( K_2Cr_2O_7 \).

### Step-by-Step Solution:

1. **Understanding Equivalent Weight**:
The equivalent weight of a substance is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n} \]
where \( n \) is the number of equivalents, also known as the n-factor.

2. **Identify the Reaction**:
In the iodometric titration, \( K_2Cr_2O_7 \) acts as an oxidizing agent. The relevant half-reaction for the reduction of dichromate ions (\( Cr_2O_7^{2-} \)) can be written as:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]
From this half-reaction, we can see that 6 electrons are involved in the reduction of 1 mole of \( Cr_2O_7^{2-} \).

3. **Determine the n-Factor**:
The n-factor for \( K_2Cr_2O_7 \) in this reaction is equal to the number of electrons transferred, which is 6. Therefore, we have:
\[ n = 6 \]

4. **Substituting in the Equivalent Weight Formula**:
Now, substituting the n-factor back into the equivalent weight formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight of } K_2Cr_2O_7}{6} \]
Thus, \( X \) in the expression \( \frac{\text{Molecular Weight}}{X} \) is equal to 6.

5. **Final Answer**:
Therefore, the value of \( X \) is:
\[ X = 6 \]