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 Fe 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, we will follow a systematic approach to find the titre value of \( K_2Cr_2O_7 \) when it reacts with \( Fe_2(SO_4)_3 \).

### Step 1: Calculate the number of moles of ferric alum.
Given:
- Mass of ferric alum = 2.41 g
- Molar mass of ferric alum = 964 g/mol

Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{2.41 \, \text{g}}{964 \, \text{g/mol}} \approx 0.0025 \, \text{mol} \]

### Step 2: Determine the milliequivalents of \( Fe_2(SO_4)_3 \).
The reaction between iron and ferric sulfate is:
\[ Fe + Fe_2(SO_4)_3 \rightarrow 3FeSO_4 \]
From the reaction, we see that 1 mole of \( Fe_2(SO_4)_3 \) reacts with 2 moles of Fe.

Thus, the number of equivalents of \( Fe_2(SO_4)_3 \) is:
\[ \text{mEq} = \text{moles} \times \text{valency} = 0.0025 \, \text{mol} \times 2 = 0.005 \, \text{equivalents} \]
Converting to milliequivalents:
\[ \text{mEq} = 0.005 \times 1000 = 5 \, \text{mEq} \]

### Step 3: Relate the milliequivalents of \( K_2Cr_2O_7 \) to \( Fe_2(SO_4)_3 \).
In acidic medium, \( K_2Cr_2O_7 \) reacts with iron as follows:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]
The n-factor for \( K_2Cr_2O_7 \) is 6, meaning 1 mole of \( K_2Cr_2O_7 \) can provide 6 equivalents.

### Step 4: Set up the equation for the titration.
Let \( V \) be the volume of \( K_2Cr_2O_7 \) solution used in mL. The molarity of \( K_2Cr_2O_7 \) is given as \( \frac{1}{60} \) M.

Using the formula for milliequivalents:
\[ \text{mEq of } K_2Cr_2O_7 = \text{molarity} \times \text{volume} \times \text{n-factor} \]
\[ 5 \, \text{mEq} = \left(\frac{1}{60} \, \text{mol/L}\right) \times V \times 6 \]

### Step 5: Solve for \( V \).
Rearranging the equation gives:
\[ V = \frac{5}{\left(\frac{1}{60}\right) \times 6} \]
Calculating:
\[ V = \frac{5 \times 60}{6} = 50 \, \text{mL} \]

### Conclusion
The titre value of \( K_2Cr_2O_7 \) when it reacts with \( Fe_2(SO_4)_3 \) is **50 mL**.

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