`Cl_(2) overset(NaOH)(rarr)NaCl+NaClO_(3)+H_(2)O`
The equivalent mass of `Cl_(2)` in the above reaction is

Options

A. M
B. `M/3`
C. `M/2`
D. `(3M)/5`

Answer

To find the equivalent mass of \( Cl_2 \) in the reaction

\[ Cl_2 + NaOH \rightarrow NaCl + NaClO_3 + H_2O, \]

we will follow these steps:

### Step 1: Determine the oxidation states of chlorine in the reactants and products.

- In \( Cl_2 \), the oxidation state of chlorine is 0.
- In \( NaCl \), the oxidation state of chlorine is -1.
- In \( NaClO_3 \), the oxidation state of chlorine can be calculated as follows:
- Let the oxidation state of \( Cl \) be \( x \).
- The oxidation state of \( Na \) is +1, and the total charge of the molecule is 0.
- Therefore, \( 1 + x + 3(-2) = 0 \) (since the oxidation state of \( O \) is -2).
- This simplifies to \( x - 5 = 0 \) or \( x = +5 \).

### Step 2: Identify the changes in oxidation states.

- Chlorine is reduced from 0 in \( Cl_2 \) to -1 in \( NaCl \) (a change of -1).
- Chlorine is oxidized from 0 in \( Cl_2 \) to +5 in \( NaClO_3 \) (a change of +5).

### Step 3: Calculate the total change in oxidation states.

- The total change in oxidation states for one mole of \( Cl_2 \) is:
- Reduction: \( -1 \) (for \( NaCl \))
- Oxidation: \( +5 \) (for \( NaClO_3 \))
- Therefore, the total change = \( 1 + 5 = 6 \).

### Step 4: Determine the number of equivalents (N).

- In a disproportionation reaction, the number of equivalents \( N \) can be calculated as:
\[ N = \frac{1}{n_1} + \frac{1}{n_2} \]
where \( n_1 \) is the change in oxidation state for reduction and \( n_2 \) is the change in oxidation state for oxidation.
- Here, \( n_1 = 1 \) and \( n_2 = 5 \).
- Thus,
\[ N = \frac{1}{1} + \frac{1}{5} = 1 + 0.2 = 1.2. \]

### Step 5: Calculate the equivalent mass of \( Cl_2 \).

- The molecular weight of \( Cl_2 \) (which is \( 2 \times 35.5 \)) is \( 71 \, g/mol \).
- The equivalent mass can be calculated using the formula:
\[ \text{Equivalent mass} = \frac{\text{Molecular weight}}{N}. \]
- Therefore,
\[ \text{Equivalent mass} = \frac{71}{1.2} \approx 59.17 \, g/equiv. \]

### Final Answer:
The equivalent mass of \( Cl_2 \) in the reaction is approximately \( 59.17 \, g/equiv \).

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