Number of electrons lost during oxidation of `0.355` g of `Cl^(-)` are :

Options

A. `0.01`
B. `0.01N_(A)`
C. `0.02N_(A)`
D. `0.01/(2N_(A))`

Answer

To find the number of electrons lost during the oxidation of 0.355 g of Cl⁻, we can follow these steps:

### Step 1: Determine the molar mass of Cl⁻
The molar mass of chlorine (Cl) is approximately 35.5 g/mol. Since we are dealing with Cl⁻, the molar mass remains the same.

### Step 2: Calculate the number of moles of Cl⁻
To find the number of moles of Cl⁻ in 0.355 g, we use the formula:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

Substituting the values:

\[ \text{Number of moles of Cl}^- = \frac{0.355 \, \text{g}}{35.5 \, \text{g/mol}} \approx 0.01 \, \text{mol} \]

### Step 3: Determine the number of electrons lost
In the oxidation process, each Cl⁻ ion loses one electron to form Cl₂. Therefore, the number of electrons lost is equal to the number of moles of Cl⁻ multiplied by Avogadro's number (approximately \(6.022 \times 10^{23}\) particles/mol):

\[ \text{Number of electrons lost} = \text{Number of moles of Cl}^- \times N_A \]

Substituting the values:

\[ \text{Number of electrons lost} = 0.01 \, \text{mol} \times 6.022 \times 10^{23} \, \text{electrons/mol} \approx 6.022 \times 10^{21} \, \text{electrons} \]

### Conclusion
The number of electrons lost during the oxidation of 0.355 g of Cl⁻ is approximately \(6.022 \times 10^{21}\) electrons.

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