The eq.wt of `K_(2)CrO_(4)` as an oxidasing agent in acid medium is

Options

A. mol.wt/2
B. `(2xx"mol.wt")/3`
C. `("mol.wt")/3`
D. `("mol.wt")/6`

Answer

To find the equivalent weight of \( K_2CrO_4 \) as an oxidizing agent in acidic medium, we can follow these steps:

### Step 1: Determine the oxidation state of chromium in \( K_2CrO_4 \)
In \( K_2CrO_4 \), the chromium is in the form of \( CrO_4^{2-} \). To find the oxidation state of chromium (Cr), we can set up the equation based on the known oxidation states of oxygen.

The oxidation state of oxygen is -2. Therefore, for \( CrO_4^{2-} \):
\[ X + 4(-2) = -2 \]
Solving for \( X \):
\[ X - 8 = -2 \\ X = +6 \]
Thus, the oxidation state of chromium in \( K_2CrO_4 \) is +6.

### Step 2: Identify the reduction process in acidic medium
In acidic medium, \( K_2CrO_4 \) acts as an oxidizing agent and gets reduced. The chromium in \( K_2CrO_4 \) (with an oxidation state of +6) is reduced to chromium with an oxidation state of +3.

### Step 3: Calculate the change in oxidation state
The change in oxidation state for chromium is from +6 to +3:
\[ \text{Change} = 6 - 3 = 3 \]
This indicates that chromium gains 3 electrons during the reduction process.

### Step 4: Determine the n-factor
The n-factor is defined as the number of electrons gained or lost per formula unit during the reaction. Since chromium gains 3 electrons, the n-factor for \( K_2CrO_4 \) as an oxidizing agent is 3.

### Step 5: Calculate the molecular weight of \( K_2CrO_4 \)
The molecular weight of \( K_2CrO_4 \) can be calculated as follows:
- Potassium (K): 39.1 g/mol (2 atoms)
- Chromium (Cr): 52.0 g/mol (1 atom)
- Oxygen (O): 16.0 g/mol (4 atoms)

Calculating the total:
\[ \text{Molecular weight} = (2 \times 39.1) + (1 \times 52.0) + (4 \times 16.0) \\ = 78.2 + 52.0 + 64.0 = 194.2 \text{ g/mol} \]

### Step 6: Calculate the equivalent weight
The equivalent weight is calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}} \]
Substituting the values we have:
\[ \text{Equivalent weight} = \frac{194.2 \text{ g/mol}}{3} = 64.7333 \text{ g/equiv} \]

### Final Answer
The equivalent weight of \( K_2CrO_4 \) as an oxidizing agent in acidic medium is approximately **64.73 g/equiv**.

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