10 g mixture of `K_(2)Cr(2)O_(7)` and `KMnO_(4)` was treated with excess of KI in acidic medium. Iodine liberated `100 cm^(3)` of 2.2 N sodium thiosulphate solution for titration. If the mass percent of `KMnO_(4)`in the mixture Z, then what is the value of 2Z/5 ?

Answer

To solve the problem, we will follow these steps:

### Step 1: Define Variables
Let the mass of \( K_2Cr_2O_7 \) in the mixture be \( x \) grams. Therefore, the mass of \( KMnO_4 \) will be \( 10 - x \) grams since the total mass of the mixture is 10 grams.

### Step 2: Determine the Equivalent Factors
- For \( KMnO_4 \):
- The oxidation state of manganese changes from +7 to +2.
- The change in oxidation state is \( 7 - 2 = 5 \).
- Therefore, the n-factor for \( KMnO_4 \) is 5.

- For \( K_2Cr_2O_7 \):
- The oxidation state of chromium changes from +6 to +3.
- The change in oxidation state is \( 6 - 3 = 3 \).
- For 2 chromium atoms, the total change is \( 2 \times 3 = 6 \).
- Therefore, the n-factor for \( K_2Cr_2O_7 \) is 6.

### Step 3: Calculate the Number of Equivalents
The number of equivalents can be calculated using the formula:
\[ \text{Number of equivalents} = \text{Number of moles} \times \text{n-factor} \]

- For \( K_2Cr_2O_7 \):
\[ \text{Number of equivalents} = \frac{x}{294} \times 6 \]

- For \( KMnO_4 \):
\[ \text{Number of equivalents} = \frac{10 - x}{158} \times 5 \]

### Step 4: Calculate the Equivalents of Sodium Thiosulphate
The sodium thiosulphate solution used for titration is given as 100 cm³ of 2.2 N. The number of equivalents of sodium thiosulphate is:
\[ \text{Number of equivalents} = \text{Normality} \times \text{Volume (in liters)} = 2.2 \times \frac{100}{1000} = 0.22 \]

### Step 5: Set Up the Equation
Now we can set up the equation based on the equivalence:
\[ \frac{6x}{294} + \frac{5(10 - x)}{158} = 0.22 \]

### Step 6: Solve the Equation
1. Simplify the equation:
\[ \frac{6x}{294} + \frac{50 - 5x}{158} = 0.22 \]

2. Multiply through by the least common multiple (LCM) of 294 and 158 to eliminate the denominators:
\[ 6x \cdot \frac{158}{294} + 50 \cdot \frac{294}{158} - 5x = 0.22 \cdot \text{LCM} \]

3. Solve for \( x \):
- After solving, we find \( x = 8.5 \) grams (mass of \( K_2Cr_2O_7 \)).
- Therefore, the mass of \( KMnO_4 \) is \( 10 - 8.5 = 1.5 \) grams.

### Step 7: Calculate Mass Percent of \( KMnO_4 \)
The mass percent of \( KMnO_4 \) in the mixture \( Z \) is calculated as:
\[ Z = \left( \frac{1.5}{10} \right) \times 100 = 15\% \]

### Step 8: Calculate \( \frac{2Z}{5} \)
Now, we calculate:
\[ \frac{2Z}{5} = \frac{2 \times 15}{5} = 6 \]

### Final Answer
The value of \( \frac{2Z}{5} \) is **6**.

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