What is the equivalent weight of `SnCl_2 + Cl_2 rarr SnCl_4`?
(mol.wt.of `SnCl_2=190`)

Options

A. 95
B. 45
C. 60
D. 30

Answer

To find the equivalent weight of the reaction `SnCl_2 + Cl_2 → SnCl_4`, we need to follow these steps:

### Step 1: Identify the Molar Mass
The molar mass of `SnCl_2` is given as 190 g/mol.

### Step 2: Determine the Change in Oxidation State
In this reaction, tin (Sn) in `SnCl_2` is oxidized to `SnCl_4`. The oxidation state of Sn in `SnCl_2` is +2, and in `SnCl_4`, it is +4. Therefore, the change in oxidation state is:
- From +2 to +4, which is an increase of 2.

### Step 3: Calculate the n-factor
The n-factor is defined as the number of electrons lost or gained per molecule in a redox reaction. Since tin loses 2 electrons during the oxidation from `SnCl_2` to `SnCl_4`, the n-factor is 2.

### Step 4: Calculate the Equivalent Weight
The equivalent weight can be calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{n-factor}} \]
Substituting the values we have:
\[ \text{Equivalent Weight} = \frac{190 \, \text{g/mol}}{2} = 95 \, \text{g/equiv} \]

### Conclusion
The equivalent weight of `SnCl_2` in the reaction `SnCl_2 + Cl_2 → SnCl_4` is 95 g/equiv.

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