`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 chlorine (Cl₂) in the reaction:

\[ \text{Cl}_2 + \text{NaOH} \rightarrow \text{NaCl} + \text{NaClO}_3 + \text{H}_2\text{O} \]

we will follow these steps:

### Step 1: Understand the Reaction
Identify the oxidation states of chlorine in the reactants and products. In Cl₂, chlorine is in the elemental state with an oxidation state of 0. In NaCl, chlorine has an oxidation state of -1, and in NaClO₃, chlorine has an oxidation state of +5.

### Step 2: Determine the Changes in Oxidation State
- Chlorine is reduced from 0 to -1 (gaining 1 electron per Cl atom).
- Chlorine is oxidized from 0 to +5 (losing 5 electrons per Cl atom).

### Step 3: Calculate the Total Change in Electrons
Since there are 2 chlorine atoms in Cl₂:
- For reduction (0 to -1):
- Each Cl atom gains 1 electron, so for 2 Cl atoms, total electrons gained = 2.

- For oxidation (0 to +5):
- Each Cl atom loses 5 electrons, so for 2 Cl atoms, total electrons lost = 10.

### Step 4: Calculate the n Factor
The n factor for a disproportionation reaction is given by:
\[ \text{n} = \frac{1}{n_1} + \frac{1}{n_2} \]
Where:
- \( n_1 \) is the number of electrons gained (2 for reduction).
- \( n_2 \) is the number of electrons lost (10 for oxidation).

Thus:
\[ \text{n} = \frac{1}{2} + \frac{1}{10} = \frac{5}{10} + \frac{1}{10} = \frac{6}{10} = \frac{3}{5} \]

### Step 5: Find the Molar Mass of Cl₂
The molar mass of Cl₂ (chlorine gas) is:
\[ \text{Molar mass of Cl}_2 = 2 \times 35.5 \, \text{g/mol} = 71 \, \text{g/mol} \]

### Step 6: Calculate the Equivalent Mass
The equivalent mass is given by the formula:
\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{\text{n factor}} \]
Substituting the values:
\[ \text{Equivalent mass} = \frac{71 \, \text{g/mol}}{\frac{3}{5}} = 71 \times \frac{5}{3} = \frac{355}{3} \approx 118.33 \, \text{g/equiv} \]

### Final Answer
Thus, the equivalent mass of Cl₂ in the reaction is:
\[ \text{Equivalent mass of Cl}_2 = \frac{355}{3} \, \text{g/equiv} \text{ or approximately } 118.33 \, \text{g/equiv} \]