The equivalent weight of `KIO_(3)` in the reaction `2Cr(OH)_(3)+4OH+KIO_(3) rarr 2CrO_(4)^(2-)+5H_(2)O+KI` is

Options

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

Explanation

To find the equivalent weight of \( KIO_3 \) in the given redox reaction, we will follow these steps:

### Step 1: Understand the Concept of Equivalent Weight
Equivalent weight is defined as the molecular weight of a substance divided by its valence factor. The valence factor is determined by the number of electrons transferred in a redox reaction.

### Step 2: Determine the Oxidation States
In \( KIO_3 \):
- The oxidation state of iodine (I) is +5.
- In the product \( KI \), the oxidation state of iodine (I) is -1.

### Step 3: Calculate the Change in Oxidation State
The change in oxidation state for iodine from \( KIO_3 \) to \( KI \) is:
\[ \text{Change} = \text{Final Oxidation State} - \text{Initial Oxidation State} = (-1) - (+5) = -6 \]
This indicates that iodine undergoes a reduction of 6 electrons.

### Step 4: Determine the Valence Factor
The valence factor for \( KIO_3 \) in this reaction is equal to the number of electrons transferred, which we calculated to be 6.

### Step 5: Calculate the Molecular Weight of \( KIO_3 \)
The molecular weight of \( KIO_3 \) can be calculated as follows:
- Potassium (K) = 39 g/mol
- Iodine (I) = 127 g/mol
- Oxygen (O) = 16 g/mol × 3 = 48 g/mol

Adding these together:
\[ \text{Molecular Weight of } KIO_3 = 39 + 127 + 48 = 214 \text{ g/mol} \]

### Step 6: Calculate the Equivalent Weight
Using the formula for equivalent weight:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{Valence Factor}} = \frac{214 \text{ g/mol}}{6} \]
Calculating this gives:
\[ \text{Equivalent Weight} = 35.67 \text{ g/equiv} \]

### Conclusion
The equivalent weight of \( KIO_3 \) in the reaction is approximately \( 35.67 \text{ g/equiv} \).

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