0.8 M `FeSO_(4)` solution requires 160mL,0.2 M `Al_(2)(Cr_(2)O_(7))_(3)` in acidic medium. Calculate volume of `FeSO_(4)` consumed :

Options

A. 480mL
B. 240mL
C. 720mL
D. 40mL

Answer

To solve the problem, we need to calculate the volume of `FeSO4` consumed in a reaction with `Al2(Cr2O7)3` in acidic medium. Let's break down the steps:

### Step 1: Understand the Reaction
In acidic medium, `Al2(Cr2O7)3` acts as an oxidizing agent. The chromium in `Cr2O7^2-` is reduced from +6 oxidation state to +3 oxidation state, while `Fe^2+` is oxidized to `Fe^3+`.

### Step 2: Calculate the Number of Moles of `Al2(Cr2O7)3`
Given:
- Concentration of `Al2(Cr2O7)3` = 0.2 M
- Volume of `Al2(Cr2O7)3` = 160 mL = 0.160 L

Using the formula:
\[ \text{Moles} = \text{Concentration} \times \text{Volume} \]
\[ \text{Moles of } Al2(Cr2O7)3 = 0.2 \, \text{mol/L} \times 0.160 \, \text{L} = 0.032 \, \text{mol} \]

### Step 3: Determine the Equivalent Factor (n-factor) of `Al2(Cr2O7)3`
In acidic medium, each `Cr2O7^2-` ion can accept 6 electrons (as it reduces from +6 to +3). Since there are 2 chromium atoms in `Cr2O7^2-`, the total change is:
\[ \text{n-factor of } Al2(Cr2O7)3 = 2 \times 6 = 12 \]
However, since there are 3 `Cr2O7^2-` ions in `Al2(Cr2O7)3`, the total n-factor is:
\[ \text{n-factor} = 12 \times 3 = 18 \]

### Step 4: Calculate the Equivalent Moles of `FeSO4`
Using the n-factor, we can calculate the equivalents of `Al2(Cr2O7)3`:
\[ \text{Equivalents of } Al2(Cr2O7)3 = \text{Moles} \times \text{n-factor} = 0.032 \, \text{mol} \times 18 = 0.576 \, \text{equivalents} \]

### Step 5: Calculate the Volume of `FeSO4` Required
The n-factor for `FeSO4` (from `Fe^2+` to `Fe^3+`) is 1. Therefore, the equivalents of `FeSO4` required will be equal to the equivalents of `Al2(Cr2O7)3`:
\[ \text{Equivalents of } FeSO4 = 0.576 \]

Now, we can calculate the volume of `FeSO4` needed:
Given:
- Concentration of `FeSO4` = 0.8 M
- n-factor of `FeSO4` = 1

Using the formula:
\[ \text{Equivalents} = \text{Moles} \times \text{n-factor} = \text{Concentration} \times \text{Volume} \]
Rearranging gives:
\[ \text{Volume} = \frac{\text{Equivalents}}{\text{Concentration}} = \frac{0.576}{0.8} = 0.72 \, \text{L} = 720 \, \text{mL} \]

### Final Answer
The volume of `FeSO4` consumed is **720 mL**.

---