Equivalent weight of `K_(2)Cr_(2)O_(4)` in acidic medium is

Options

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

Explanation

To find the equivalent weight of \( K_2Cr_2O_4 \) (potassium dichromate) in acidic medium, we can follow these steps:

### Step 1: Determine the oxidation states of chromium
In \( K_2Cr_2O_4 \), the oxidation state of chromium (Cr) is +6. When it reacts in acidic medium, it is reduced to +3.

### Step 2: Calculate the change in oxidation state
The change in oxidation state for each chromium atom is:
\[ +6 \text{ (in } K_2Cr_2O_4\text{)} \rightarrow +3 \text{ (in } Cr^{3+}\text{)} \]
The decrease in oxidation state per chromium atom is:
\[ 6 - 3 = 3 \]
Since there are 2 chromium atoms in \( K_2Cr_2O_4 \), the total decrease in oxidation state is:
\[ 2 \times 3 = 6 \]

### Step 3: Use the formula for equivalent weight
The equivalent weight (EW) is calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n}} \]
where \( n \) is the number of electrons transferred per formula unit in the reaction. Here, \( n = 6 \) (the total decrease in oxidation state).

### Step 4: Calculate the molecular weight of \( K_2Cr_2O_4 \)
The molecular weight of \( K_2Cr_2O_4 \) can be calculated as follows:
- Potassium (K): 39.1 g/mol (2 atoms)
- Chromium (Cr): 52.0 g/mol (2 atoms)
- Oxygen (O): 16.0 g/mol (4 atoms)

Calculating the total:
\[ \text{Molecular weight} = (2 \times 39.1) + (2 \times 52.0) + (4 \times 16.0) = 78.2 + 104.0 + 64.0 = 246.2 \text{ g/mol} \]

### Step 5: Calculate the equivalent weight
Now, substituting the values into the equivalent weight formula:
\[ \text{Equivalent weight} = \frac{246.2 \text{ g/mol}}{6} \approx 41.03 \text{ g/equiv} \]

### Final Answer
The equivalent weight of \( K_2Cr_2O_4 \) in acidic medium is approximately \( 41.03 \text{ g/equiv} \).

---