An acidic solution of `Cu^(2+)` ions containing 0.4 g of `Cu^(2+)` ions is electrolysed until all the copper is deposited. Calculate the volume of oxygen evolved at N.T.P.

Options

A. 141 cc
B. 31.75 cc
C. 64 cc
D. 32 cc

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of `Cu^(2+)` ions
To find the number of moles of `Cu^(2+)` ions in 0.4 g, we need to use the molar mass of copper.

- Molar mass of Cu = 63.5 g/mol
- Number of moles of Cu = mass (g) / molar mass (g/mol)

\[ \text{Number of moles of } Cu^{2+} = \frac{0.4 \, \text{g}}{63.5 \, \text{g/mol}} \approx 0.0063 \, \text{mol} \]

### Step 2: Determine the equivalent weight of `Cu^(2+)`
The equivalent weight of copper can be calculated using the formula:

\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{Valency factor}} \]

For `Cu^(2+)`, the valency factor is 2 (since it gains 2 electrons).

\[ \text{Equivalent weight of } Cu^{2+} = \frac{63.5 \, \text{g/mol}}{2} = 31.75 \, \text{g/equiv} \]

### Step 3: Calculate the equivalents of `Cu^(2+)`
Now we can find the number of equivalents of `Cu^(2+)` in the solution.

\[ \text{Equivalents of } Cu^{2+} = \frac{\text{mass of } Cu^{2+}}{\text{Equivalent weight}} = \frac{0.4 \, \text{g}}{31.75 \, \text{g/equiv}} \approx 0.0126 \, \text{equiv} \]

### Step 4: Relate the equivalents of `Cu^(2+)` to the volume of `O_2` evolved
From the electrolysis reaction, we know that 1 equivalent of `Cu^(2+)` produces 0.5 equivalents of `O_2` gas. Therefore, the equivalents of `O_2` produced will be:

\[ \text{Equivalents of } O_2 = 0.0126 \, \text{equiv} \times 0.5 = 0.0063 \, \text{equiv} \]

### Step 5: Calculate the volume of `O_2` at NTP
At NTP (Normal Temperature and Pressure), 1 equivalent of gas occupies 22.4 L. Therefore, the volume of `O_2` produced is:

\[ \text{Volume of } O_2 = \text{Equivalents of } O_2 \times 22.4 \, \text{L/equiv } = 0.0063 \, \text{equiv} \times 22.4 \, \text{L/equiv} \approx 0.141 \, \text{L} \]

### Step 6: Convert the volume from liters to cubic centimeters
Since 1 L = 1000 cm³, we convert the volume:

\[ \text{Volume of } O_2 = 0.141 \, \text{L} \times 1000 \, \text{cm}^3/\text{L} = 141 \, \text{cm}^3 \]

### Final Answer
The volume of oxygen evolved at NTP is **141 cm³**.

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