`Cl_(2)` changes to` Cl^(-)` in cold NaOH. The equivalent weight of `Cl_(2)` will be :

Explanation

To find the equivalent weight of \( Cl_2 \) when it reacts with cold \( NaOH \), we can follow these steps:

### Step 1: Write the reaction
The reaction of chlorine gas (\( Cl_2 \)) with sodium hydroxide (\( NaOH \)) in cold conditions can be represented as:
\[ Cl_2 + 2NaOH \rightarrow NaCl + NaOCl + H_2O \]

### Step 2: Balance the reaction
In this reaction, we have:
- 1 molecule of \( Cl_2 \)
- 2 molecules of \( NaOH \)
- Producing 1 molecule of \( NaCl \), 1 molecule of \( NaOCl \), and 1 molecule of water (\( H_2O \))

The reaction is already balanced as written.

### Step 3: Determine the n-factor
The n-factor of a substance in a reaction is defined as the number of moles of electrons transferred per mole of the substance. In this case:
- \( Cl_2 \) is reduced to \( Cl^- \) in \( NaCl \) and \( ClO^- \) in \( NaOCl \).
- Each \( Cl \) atom in \( Cl_2 \) is reduced by 1 electron to form \( Cl^- \) and contributes to the formation of \( ClO^- \).

Thus, the n-factor for \( Cl_2 \) is 1 (since one mole of \( Cl_2 \) gives one mole of \( Cl^- \)).

### Step 4: Calculate the molecular weight of \( Cl_2 \)
The molecular weight of \( Cl_2 \) can be calculated as follows:
- The atomic weight of chlorine (Cl) is approximately 35.5 g/mol.
- Therefore, the molecular weight of \( Cl_2 \) is:
\[ Molecular \ weight \ of \ Cl_2 = 2 \times 35.5 = 71 \ g/mol \]

### Step 5: Calculate the equivalent weight
The equivalent weight is calculated using the formula:
\[ Equivalent \ weight = \frac{Molecular \ weight}{n-factor} \]
Substituting the values we have:
\[ Equivalent \ weight = \frac{71 \ g/mol}{1} = 71 \ g/equiv \]

### Final Answer
Thus, the equivalent weight of \( Cl_2 \) is **71 g/equiv**.

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