`Z_(2)` undergo disproportion in an alkaline medium into a mixture of `Z^(-)` and `ZO_(3)^(-)`. Equivalent mass of `Z_(2)` in the reaction is: [given that atomic mass of Z is 80 gm/mol]

Options

A. 96
B. 48
C. 64
D. 32

Explanation

To find the equivalent mass of \( Z_2 \) in the given disproportionation reaction, we can follow these steps:

### Step 1: Write the Reaction
The reaction involves \( Z_2 \) undergoing disproportionation in an alkaline medium to form \( Z^- \) and \( ZO_3^- \). The reaction can be represented as:
\[ Z_2 \rightarrow Z^- + ZO_3^- \]

### Step 2: Determine Oxidation States
- In \( Z_2 \), the oxidation state of \( Z \) is 0.
- In \( Z^- \), the oxidation state of \( Z \) is -1.
- In \( ZO_3^- \), we need to find the oxidation state of \( Z \). The oxidation state of oxygen is -2. Let the oxidation state of \( Z \) be \( x \):
\[ x + 3(-2) = -1 \implies x - 6 = -1 \implies x = +5 \]
Thus, the oxidation state of \( Z \) in \( ZO_3^- \) is +5.

### Step 3: Identify the Changes in Oxidation States
- For the reduction from \( Z_2 \) to \( Z^- \):
- Change in oxidation state: \( 0 \) to \( -1 \) (gain of 1 electron per atom of \( Z \)).
- For \( Z_2 \) (2 atoms), total electrons gained = \( 2 \).

- For the oxidation from \( Z_2 \) to \( ZO_3^- \):
- Change in oxidation state: \( 0 \) to \( +5 \) (loss of 5 electrons per atom of \( Z \)).
- For \( Z_2 \) (2 atoms), total electrons lost = \( 2 \times 5 = 10 \).

### Step 4: Calculate the n-factor
The n-factor is the total number of electrons exchanged in the reaction:
\[ \text{n-factor} = \text{electrons gained} + \text{electrons lost} = 2 + 10 = 12 \]

### Step 5: Calculate the Equivalent Mass
The equivalent mass is given by the formula:
\[ \text{Equivalent mass} = \frac{\text{Molecular mass}}{\text{n-factor}} \]
The molecular mass of \( Z_2 \) (where the atomic mass of \( Z \) is 80 g/mol) is:
\[ \text{Molecular mass of } Z_2 = 2 \times 80 = 160 \text{ g/mol} \]
Now, substituting the values:
\[ \text{Equivalent mass} = \frac{160 \text{ g/mol}}{12} = \frac{160}{12} \approx 13.33 \text{ g/equiv} \]

### Final Answer
The equivalent mass of \( Z_2 \) in the reaction is approximately **13.33 g/equiv**.

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