Same quantity of current is passed through molten `NaCl` and molten cryolite containing `Al_(2)O_(3)`. If `4.6g` of `Na` was liberated in one cell, the mass of `Al` liberated in the other cell is
1) `0.9g`
2) `1.8g`
3) `2.7g`
4) `3.6g`

Options

A. `0.9g`
B. `1.8g`
C. `2.7g`
D. `3.6g`

Explanation

To solve the problem, we will use Faraday's laws of electrolysis, particularly Faraday's third law, which states that when the same quantity of electricity is passed through different electrolytic cells, the mass of substances liberated at the electrodes is directly proportional to their equivalent weights.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of Sodium (Na) liberated, \( W_1 = 4.6 \, \text{g} \)
- Equivalent weight of Sodium, \( E_1 = \frac{M}{n} = \frac{23 \, \text{g/mol}}{1} = 23 \, \text{g/equiv} \)
- Equivalent weight of Aluminium (Al), \( E_2 = \frac{M}{n} = \frac{27 \, \text{g/mol}}{3} = 9 \, \text{g/equiv} \)

2. **Apply Faraday's Third Law:**
According to Faraday's third law:
\[ \frac{W_1}{E_1} = \frac{W_2}{E_2} \]
where \( W_2 \) is the mass of Aluminium liberated.

3. **Rearranging the Equation:**
We can rearrange the equation to solve for \( W_2 \):
\[ W_2 = W_1 \times \frac{E_2}{E_1} \]

4. **Substituting the Values:**
Substitute the known values into the equation:
\[ W_2 = 4.6 \, \text{g} \times \frac{9 \, \text{g/equiv}}{23 \, \text{g/equiv}} \]

5. **Calculating \( W_2 \):**
\[ W_2 = 4.6 \times \frac{9}{23} = 4.6 \times 0.3913 \approx 1.8 \, \text{g} \]

6. **Final Result:**
The mass of Aluminium liberated in the other cell is \( W_2 \approx 1.8 \, \text{g} \).

### Conclusion:
Thus, the answer is \( \text{Option 2: } 1.8 \, \text{g} \).