The equivalent weight of salt
`KHC_(2)O_(4).H_(2)C_(2)O_(4).4H_(2)O` when used as reducing agent `:-`

Options

A. `("Mol.mass")/(1)`
B. `("Mol.mass")/(2)`
C. `("Mol.mass")/(3)`
D. `("Mol.mass")/(4)`

Answer

To find the equivalent weight of the salt \( KHC_2O_4 \cdot H_2C_2O_4 \cdot 4H_2O \) when used as a reducing agent, we can follow these steps:

### Step 1: Determine the Molecular Weight of the Salt

The molecular formula of the salt is \( KHC_2O_4 \cdot H_2C_2O_4 \cdot 4H_2O \). We need to calculate its molar mass by adding the atomic weights of all the atoms in the formula.

- Potassium (K): 39.1 g/mol
- Hydrogen (H): 1.0 g/mol (5 H in total)
- Carbon (C): 12.0 g/mol (4 C in total)
- Oxygen (O): 16.0 g/mol (8 O in total)

Calculating the total:
\[ \text{Molar mass} = 39.1 + (5 \times 1.0) + (4 \times 12.0) + (8 \times 16.0) \]
\[ = 39.1 + 5 + 48 + 128 = 220.1 \text{ g/mol} \]

### Step 2: Determine the Change in Oxidation State

Next, we need to find the change in oxidation state of carbon when the salt acts as a reducing agent.

In \( KHC_2O_4 \) and \( H_2C_2O_4 \), the oxidation state of carbon is +3. When it acts as a reducing agent, it gets oxidized to +4.

The change in oxidation state for one carbon atom is:
\[ \Delta \text{Oxidation State} = +4 - (+3) = +1 \]

Since there are 4 carbon atoms in total (2 from \( KHC_2O_4 \) and 2 from \( H_2C_2O_4 \)), the total change in oxidation state is:
\[ \text{Total Change} = 1 \times 4 = 4 \]

### Step 3: Determine the n-factor

The n-factor (valency factor) for the salt when it acts as a reducing agent is equal to the total change in oxidation state, which we calculated to be 4.

### Step 4: Calculate the Equivalent Weight

The equivalent weight of a compound is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{n\text{-factor}} \]

Substituting the values we have:
\[ \text{Equivalent Weight} = \frac{220.1 \text{ g/mol}}{4} = 55.025 \text{ g/equiv} \]

### Conclusion

The equivalent weight of the salt \( KHC_2O_4 \cdot H_2C_2O_4 \cdot 4H_2O \) when used as a reducing agent is approximately **55.025 g/equiv**.

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