The weight of oxalic acid required to reduce 80 ml of 0.4 M `KMnO_(4)` in acidic medium is

Options

A. 10.08 g
B. 7.21 g
C. 16.28 g
D. 12.4 g

Explanation

To solve the problem of determining the weight of oxalic acid required to reduce 80 ml of 0.4 M KMnO4 in acidic medium, we can follow these steps:

### Step 1: Calculate the number of moles of KMnO4
We start by calculating the number of moles of KMnO4 present in the solution.

\[ \text{Moles of KMnO}_4 = \text{Concentration} \times \text{Volume} \]

Given:
- Concentration of KMnO4 = 0.4 M
- Volume of KMnO4 = 80 ml = 0.080 L (since 1 L = 1000 ml)

\[ \text{Moles of KMnO}_4 = 0.4 \, \text{mol/L} \times 0.080 \, \text{L} = 0.032 \, \text{mol} \]

### Step 2: Determine the n-factor for KMnO4
In acidic medium, KMnO4 (MnO4^-) is reduced from +7 oxidation state to +2 oxidation state (Mn^2+). The change in oxidation state is:

\[ \text{Change in oxidation state} = +7 \text{ to } +2 = 5 \]

Since there is 1 manganese atom undergoing this change, the n-factor for KMnO4 is 5.

### Step 3: Calculate the equivalents of KMnO4
The number of equivalents of KMnO4 can be calculated using the formula:

\[ \text{Equivalents of KMnO}_4 = \text{Moles} \times \text{n-factor} \]

\[ \text{Equivalents of KMnO}_4 = 0.032 \, \text{mol} \times 5 = 0.16 \, \text{equivalents} \]

### Step 4: Determine the n-factor for oxalic acid
Oxalic acid (H2C2O4) is oxidized in the reaction. The oxidation state of carbon in oxalic acid changes from +3 to +4. The change in oxidation state for each carbon atom is 1, and since there are 2 carbon atoms, the n-factor for oxalic acid is:

\[ \text{n-factor of oxalic acid} = 1 \times 2 = 2 \]

### Step 5: Calculate the equivalents of oxalic acid required
Since the equivalents of KMnO4 and oxalic acid must be equal in a redox reaction, we have:

\[ \text{Equivalents of oxalic acid} = \text{Equivalents of KMnO}_4 = 0.16 \, \text{equivalents} \]

### Step 6: Calculate the number of moles of oxalic acid
Using the n-factor of oxalic acid, we can calculate the moles of oxalic acid required:

\[ \text{Moles of oxalic acid} = \frac{\text{Equivalents}}{\text{n-factor}} = \frac{0.16}{2} = 0.08 \, \text{mol} \]

### Step 7: Calculate the weight of oxalic acid
To find the weight of oxalic acid, we use the formula:

\[ \text{Weight} = \text{Moles} \times \text{Molar mass} \]

The molar mass of oxalic acid (H2C2O4) is approximately 90 g/mol.

\[ \text{Weight} = 0.08 \, \text{mol} \times 90 \, \text{g/mol} = 7.2 \, \text{g} \]

### Final Answer
The weight of oxalic acid required to reduce 80 ml of 0.4 M KMnO4 in acidic medium is **7.2 grams**.

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