100 mL solution of ferric alum `[Fe_(2)(SO_(4))_(3).(NH_(4))_(2)SO_(4).24H_(2)O`
`(Mw=964g mol^(-1))` containing `2.41 g ` salt was boiled with Fe when the reaction
`Fe+Fe_(2)(SO_(4))_(3)to3FeSO_(4)`
Takes place. The unreacted iron was filtered off and the solution was titrated with `(M)/(60)K_(2)Cr_(2)O_(7)` in acidic medium.
Q. What is the titre value of `K_(2)Cr_(2)O_(7)` when Cu reacts with `Fe_(2)(SO_(4))_(3)`?

Options

A. 25 mL
B. 50 mL
C. 75 mL
D. 100 ml

Explanation

To solve the problem step by step, we need to determine the titre value of \( K_2Cr_2O_7 \) when it reacts with \( Fe_2(SO_4)_3 \). We will follow the stoichiometric calculations based on the information provided.

### Step 1: Calculate the number of moles of ferric alum in the solution.

Given:
- Mass of ferric alum = 2.41 g
- Molar mass of ferric alum \( (Fe_2(SO_4)_3 \cdot (NH_4)_2SO_4 \cdot 24H_2O) = 964 \, g/mol \)

\[ \text{Number of moles of ferric alum} = \frac{\text{mass}}{\text{molar mass}} = \frac{2.41 \, g}{964 \, g/mol} = 0.0025 \, mol \]

### Step 2: Determine the equivalents of \( Fe_2(SO_4)_3 \).

The reaction between iron and ferric alum can be represented as:
\[ Fe + Fe_2(SO_4)_3 \rightarrow 3FeSO_4 \]

From the reaction, we see that 1 mole of \( Fe_2(SO_4)_3 \) reacts with 2 moles of Fe. Therefore, the number of equivalents of \( Fe_2(SO_4)_3 \) is equal to the number of moles multiplied by 3 (since it produces 3 moles of \( FeSO_4 \)).

\[ \text{Equivalents of } Fe_2(SO_4)_3 = 0.0025 \, mol \times 3 = 0.0075 \, eq \]

### Step 3: Relate the equivalents of \( K_2Cr_2O_7 \) to the equivalents of \( FeSO_4 \).

In acidic medium, \( K_2Cr_2O_7 \) has an n-factor of 6 because it undergoes a reduction from \( Cr_2O_7^{2-} \) (oxidation state +6) to \( Cr^{3+} \) (oxidation state +3).

### Step 4: Calculate the milliequivalents of \( K_2Cr_2O_7 \).

Given the strength of \( K_2Cr_2O_7 \) is \( \frac{M}{60} \):
\[ \text{Milliequivalents of } K_2Cr_2O_7 = \text{Equivalents of } FeSO_4 \]

From the previous steps:
\[ 0.0075 \, eq = \frac{M}{60} \times V \times 6 \]

### Step 5: Solve for volume \( V \).

Rearranging the equation to solve for \( V \):
\[ V = \frac{0.0075 \, eq}{\frac{M}{60} \times 6} \]

Substituting \( M = 1 \, mol/L \):
\[ V = \frac{0.0075}{\frac{1}{60} \times 6} = \frac{0.0075 \times 60}{6} = 0.075 \, L = 75 \, mL \]

### Conclusion

The titre value of \( K_2Cr_2O_7 \) when it reacts with \( Fe_2(SO_4)_3 \) is 75 mL.

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