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

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, which relate the amount of substance liberated during electrolysis to the quantity of electricity passed through the electrolyte.

### Step-by-Step Solution:

1. **Identify the Electrochemical Reactions**:
- In the electrolysis of molten NaCl, sodium (Na) is liberated at the cathode:
\[ \text{Na}^+ + e^- \rightarrow \text{Na} \]
- In the electrolysis of molten cryolite (which contains Al₂O₃), aluminum (Al) is liberated at the cathode:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]

2. **Calculate the Moles of Sodium (Na) Liberated**:
- The molar mass of sodium (Na) is approximately 23 g/mol.
- Given that 4.6 g of Na is liberated, we can calculate the number of moles of Na:
\[ \text{Moles of Na} = \frac{4.6 \text{ g}}{23 \text{ g/mol}} \approx 0.2 \text{ moles} \]

3. **Determine the Charge Used for Sodium Liberation**:
- Each mole of Na requires 1 mole of electrons (1 Faraday) for its reduction. Therefore, the total charge (Q) used to liberate Na can be calculated using Faraday's constant (approximately 96500 C/mol):
\[ Q = \text{Moles of Na} \times \text{Faraday's constant} = 0.2 \text{ moles} \times 96500 \text{ C/mol} \approx 19300 \text{ C} \]

4. **Relate Charge to Aluminum Liberation**:
- For aluminum, 1 mole of Al requires 3 moles of electrons for its reduction. Therefore, the charge required to liberate 1 mole of Al is:
\[ Q_{\text{Al}} = 3 \times \text{Faraday's constant} = 3 \times 96500 \text{ C/mol} \approx 289500 \text{ C} \]

5. **Calculate the Moles of Aluminum (Al) Liberated**:
- Since the same quantity of current is passed through both cells, we can set up a proportion based on the charge:
\[ Q_{\text{Na}} = Q_{\text{Al}} \times \text{Moles of Al} \]
- Rearranging gives:
\[ \text{Moles of Al} = \frac{Q_{\text{Na}}}{Q_{\text{Al}}} = \frac{19300 \text{ C}}{289500 \text{ C/mol}} \approx 0.0667 \text{ moles} \]

6. **Calculate the Mass of Aluminum (Al) Liberated**:
- The molar mass of aluminum (Al) is approximately 27 g/mol. Thus, the mass of Al liberated can be calculated as:
\[ \text{Mass of Al} = \text{Moles of Al} \times \text{Molar mass of Al} = 0.0667 \text{ moles} \times 27 \text{ g/mol} \approx 1.80 \text{ g} \]

### Final Answer:
The mass of aluminum (Al) liberated in the other cell is approximately **1.80 g**.