0.5 gmixture of `K_(2)Cr_(2)O_(7)` and `KMnO_(4)` was treated with excess of KI in acidic medium. Iodine liberated required `150 cm^(3)` of 0.10 N solution of thiosulphate solution for titration.
Find trhe percentage of `K_(2)Cr_(2)O_(7)` in the mixture :

Options

A. 14.64
B. 34.2
C. 65.69
D. 50

Answer

To solve the problem, we need to determine the percentage of \( K_2Cr_2O_7 \) in a mixture of \( K_2Cr_2O_7 \) and \( KMnO_4 \) that weighs 0.5 g. The iodine liberated from the reaction with excess KI in an acidic medium is titrated with a thiosulfate solution.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction involves \( K_2Cr_2O_7 \) and \( KMnO_4 \) reacting with \( KI \) in acidic medium, which liberates iodine (\( I_2 \)). The relevant half-reactions are:
- \( K_2Cr_2O_7 \) reduces \( Cr^{6+} \) to \( Cr^{3+} \)
- \( KMnO_4 \) reduces \( Mn^{7+} \) to \( Mn^{2+} \)
- \( I^- \) is oxidized to \( I_2 \)

2. **Determine the N-Factors**:
- For \( K_2Cr_2O_7 \): The change in oxidation state for chromium is \( 6 \) (from +6 to +3). Since there are 2 chromium atoms, the n-factor is \( 6 \).
- For \( KMnO_4 \): The change in oxidation state for manganese is \( 5 \) (from +7 to +2). Thus, the n-factor is \( 5 \).
- For \( I^- \): The change in oxidation state is \( 1 \) (from -1 to 0). Thus, the n-factor is \( 1 \).

3. **Set Up the Equations**:
Let \( x \) be the mass of \( KMnO_4 \) and \( y \) be the mass of \( K_2Cr_2O_7 \). We know:
\[ x + y = 0.5 \quad \text{(1)} \]

The equivalents of iodine liberated can be expressed as:
\[ \text{Equivalents of } I_2 = \frac{5x}{158} + \frac{6y}{294} \]

This is equal to the equivalents of thiosulfate used:
\[ \text{Equivalents of thiosulfate} = \text{Normality} \times \text{Volume} = 0.1 \times \frac{150}{1000} = 0.015 \quad \text{(2)} \]

4. **Combine the Equations**:
From (1) and (2), we can set up:
\[ \frac{5x}{158} + \frac{6y}{294} = 0.015 \]

5. **Substitute \( y \) in Terms of \( x \)**:
From equation (1), we can express \( y \):
\[ y = 0.5 - x \]

Substitute \( y \) into the equivalents equation:
\[ \frac{5x}{158} + \frac{6(0.5 - x)}{294} = 0.015 \]

6. **Solve for \( x \)**:
Simplifying the equation:
\[ \frac{5x}{158} + \frac{3 - 6x}{294} = 0.015 \]

To solve for \( x \), find a common denominator and solve the resulting equation.

7. **Calculate \( y \)**:
Once \( x \) is found, substitute back to find \( y \).

8. **Calculate the Percentage of \( K_2Cr_2O_7 \)**:
Finally, the percentage of \( K_2Cr_2O_7 \) in the mixture is:
\[ \text{Percentage of } K_2Cr_2O_7 = \left( \frac{y}{0.5} \right) \times 100 \]

### Final Calculation:
After calculations, we find:
- \( y \) (mass of \( K_2Cr_2O_7 \)) = 0.073 g
- Percentage of \( K_2Cr_2O_7 \) = \( \left( \frac{0.073}{0.5} \right) \times 100 = 14.6\% \)