When molten magnesium oxide was electrolysed for a certain period , 150 mg of Mg was deposited on the cathode . The volume of oxygen gas in `cm^3` at STP conditions liberated at the anode during the same period is (Atomic mass of Mg = 24 `g mol^(-1)`)

Options

A. 140
B. 280
C. 70
D. 120

Explanation

To solve the problem of determining the volume of oxygen gas liberated at the anode during the electrolysis of molten magnesium oxide, we can follow these steps:

### Step-by-Step Solution:

1. **Convert the mass of magnesium deposited to grams**:
\[ \text{Mass of Mg} = 150 \, \text{mg} = 150 \times 10^{-3} \, \text{g} = 0.150 \, \text{g} \]

2. **Calculate the number of moles of magnesium deposited**:
The molar mass of magnesium (Mg) is given as 24 g/mol. The number of moles (n) can be calculated using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.150 \, \text{g}}{24 \, \text{g/mol}} = 0.00625 \, \text{mol} \]

3. **Determine the equivalent weight of magnesium**:
Magnesium (Mg) has a valency of 2, meaning it requires 2 electrons to deposit 1 mole of magnesium. Therefore, the equivalent weight (E) is:
\[ E_{\text{Mg}} = \frac{\text{Molar mass}}{\text{Valency}} = \frac{24 \, \text{g/mol}}{2} = 12 \, \text{g/equiv} \]

4. **Calculate the equivalent weight of oxygen**:
Oxygen (O2) has a molar mass of 32 g/mol and requires 4 electrons to produce 1 mole of O2 (since each O atom requires 2 electrons). Therefore, the equivalent weight of oxygen is:
\[ E_{\text{O2}} = \frac{\text{Molar mass}}{\text{Valency}} = \frac{32 \, \text{g/mol}}{4} = 8 \, \text{g/equiv} \]

5. **Use the first law of electrolysis to find the mass of oxygen liberated**:
According to the first law of electrolysis:
\[ \frac{W_{\text{Mg}}}{W_{\text{O2}}} = \frac{E_{\text{Mg}}}{E_{\text{O2}}} \]
Where:
- \(W_{\text{Mg}} = 0.150 \, \text{g}\)
- \(E_{\text{Mg}} = 12 \, \text{g/equiv}\)
- \(E_{\text{O2}} = 8 \, \text{g/equiv}\)

Rearranging gives:
\[ W_{\text{O2}} = W_{\text{Mg}} \times \frac{E_{\text{O2}}}{E_{\text{Mg}}} \]
Substituting the values:
\[ W_{\text{O2}} = 0.150 \, \text{g} \times \frac{8}{12} = 0.100 \, \text{g} \]

6. **Calculate the number of moles of oxygen gas liberated**:
\[ n_{\text{O2}} = \frac{W_{\text{O2}}}{\text{Molar mass of O2}} = \frac{0.100 \, \text{g}}{32 \, \text{g/mol}} = 0.003125 \, \text{mol} \]

7. **Calculate the volume of oxygen gas at STP**:
At standard temperature and pressure (STP), 1 mole of gas occupies 22,400 cm³. Therefore, the volume of oxygen gas can be calculated as:
\[ \text{Volume} = n_{\text{O2}} \times 22,400 \, \text{cm}^3/\text{mol} = 0.003125 \, \text{mol} \times 22,400 \, \text{cm}^3/\text{mol} = 70 \, \text{cm}^3 \]

### Final Answer:
The volume of oxygen gas liberated at the anode during the electrolysis is **70 cm³**.