During electrolysis, two cells each containing molten Al\(_2\)O\(_3\) and fuse CaCl\(_2\) were connected in series....

Explanation

During electrolysis, the amount of substance deposited at the cathode is directly proportional to the amount of electric charge passed through the electrolyte. To find the mass of aluminum deposited, we can use the concept of Faraday's laws of electrolysis.

First, let's determine the Faraday's constant (F) which is the electric charge per mole of electrons:

\[F = 96500\,\text{C/mol}\]

Next, we need to determine the number of moles of calcium (n) that were deposited:

\[n_\text{Ca} = \frac{\text{mass of Ca}}{\text{molar mass of Ca}} = \frac{9\,\text{g}}{40\,\text{g/mol}} = 0.225\,\text{mol}\]

Now, let's calculate the number of Faraday's (F) used to deposit this number of moles of calcium:

\[F_\text{Ca} = n_\text{Ca} \times F = 0.225\,\text{mol} \times 96500\,\text{C/mol} = 21712.5\,\text{C}\]

Since the same current was passed through both cells, the same amount of charge was used for the deposition of aluminum. We can determine the number of moles of aluminum (n) deposited by dividing the charge used by the Faraday's constant:

\[n_\text{Al} = \frac{F_\text{Al}}{F} = \frac{21712.5\,\text{C}}{96500\,\text{C/mol}} = 0.225\,\text{mol}\]

Note that the molar ratio of calcium to aluminum deposited is 2:3. So, we need to multiply the number of moles of aluminum by 3/2 to account for this ratio:

\[n_\text{Al}^\text{corrected} = n_\text{Al} \times \frac{3}{2} = 0.225\,\text{mol} \times \frac{3}{2} = 0.3375\,\text{mol}\]

Finally, we can find the mass of aluminum deposited by multiplying the corrected number of moles by the molar mass of aluminum:

\[\text{mass of Al} = n_\text{Al}^\text{corrected} \times \text{molar mass of Al} = 0.3375\,\text{mol} \times 27\,\text{g/mol} \approx 4.05\,\text{g}\]

So, the mass of aluminum deposited in the second cell is approximately 4.05g, which corresponds to Option D.