`6xx10^(-3)` mole `K_(2)Cr_(2) C_(7)` reacts completely with `9xx10^(-3)` mole `X^(n+)` to give `XO_(3)^(-)` and `Cr^(3+)`. The value of `n` is `:`

Options

A. 1
B. 2
C. 3
D. none of these

Answer

To solve the problem, we need to determine the value of \( n \) in the reaction between potassium dichromate (\( K_2Cr_2O_7 \)) and \( X^{n+} \) that produces \( XO_3^{-} \) and \( Cr^{3+} \). Here’s a step-by-step breakdown of the solution:

### Step 1: Write the balanced chemical equation
The reaction can be represented as follows:
\[ K_2Cr_2O_7 + X^{n+} \rightarrow Cr^{3+} + XO_3^{-} \]

### Step 2: Determine the change in oxidation state for chromium
In \( K_2Cr_2O_7 \), the oxidation state of chromium (\( Cr \)) is +6. When it is reduced to \( Cr^{3+} \), it gains 3 electrons per chromium atom. Since there are 2 chromium atoms, the total number of electrons gained is:
\[ 2 \times 3 = 6 \text{ electrons} \]
Thus, the n-factor for \( K_2Cr_2O_7 \) is 6.

### Step 3: Determine the change in oxidation state for \( X^{n+} \)
For \( XO_3^{-} \), we can find the oxidation state of \( X \). The total charge of \( XO_3^{-} \) is -1, and with three oxygen atoms each having an oxidation state of -2, we can set up the equation:
\[ n + 3(-2) = -1 \implies n - 6 = -1 \implies n = +5 \]
Thus, the oxidation state of \( X \) changes from \( n \) to +5. The change in electrons for one \( X \) atom is:
\[ 5 - n \]
So, the n-factor for \( X^{n+} \) is \( 5 - n \).

### Step 4: Set up the equivalence equation
Since the number of equivalents of \( K_2Cr_2O_7 \) must equal the number of equivalents of \( X^{n+} \), we can express this as:
\[ \text{Number of equivalents of } K_2Cr_2O_7 = \text{Number of equivalents of } X^{n+} \]
This can be expressed mathematically as:
\[ \text{moles of } K_2Cr_2O_7 \times \text{n-factor of } K_2Cr_2O_7 = \text{moles of } X^{n+} \times \text{n-factor of } X^{n+} \]

### Step 5: Substitute the values
We know:
- Moles of \( K_2Cr_2O_7 = 6 \times 10^{-3} \)
- Moles of \( X^{n+} = 9 \times 10^{-3} \)
- n-factor of \( K_2Cr_2O_7 = 6 \)
- n-factor of \( X^{n+} = 5 - n \)

Substituting these values into the equation gives:
\[ (6 \times 10^{-3}) \times 6 = (9 \times 10^{-3}) \times (5 - n) \]

### Step 6: Simplify and solve for \( n \)
Simplifying the left side:
\[ 36 \times 10^{-3} = (9 \times 10^{-3}) \times (5 - n) \]
Dividing both sides by \( 9 \times 10^{-3} \):
\[ 4 = 5 - n \]
Rearranging gives:
\[ n = 5 - 4 = 1 \]

### Conclusion
Thus, the value of \( n \) is \( 1 \).