In the Hall process, aluminium is produced by the electrolysis of molten `Al_2O_3`. How many second would it take to produce enough aluminium by the Hall process to make a case of 24 cans of auminium soft-drink, if each can uses 5.0g of Al, a current of 9650amp is employed and the current efficiency of the cell is 90%:

Options

A. 203.2
B. 148.14
C. 333
D. 6.17

Answer

To solve the problem of how long it would take to produce enough aluminum for a case of 24 cans using the Hall process, we will follow these steps:

### Step 1: Calculate the total mass of aluminum needed
Each can uses 5.0 grams of aluminum, and there are 24 cans.

\[ \text{Total mass of Al} = \text{mass per can} \times \text{number of cans} = 5.0 \, \text{g} \times 24 = 120 \, \text{g} \]

### Step 2: Determine the equivalent weight of aluminum
The equivalent weight (E) of aluminum can be calculated using its atomic weight and its n-factor. The atomic weight of aluminum (Al) is approximately 27 g/mol, and in Al₂O₃, aluminum has a valency of +3.

\[ \text{n-factor} = 3 \quad \text{(since Al is in +3 oxidation state)} \]
\[ \text{Equivalent weight} = \frac{\text{Atomic weight}}{\text{n-factor}} = \frac{27 \, \text{g/mol}}{3} = 9 \, \text{g/equiv} \]

### Step 3: Calculate the charge required to produce the aluminum
Using Faraday's first law of electrolysis, the charge (Q) required can be calculated as follows:

\[ Q = \frac{\text{Weight} \times 96500}{\text{Equivalent weight}} \]

Substituting the values:

\[ Q = \frac{120 \, \text{g} \times 96500}{9 \, \text{g/equiv}} = \frac{11580000}{9} \approx 1286666.67 \, \text{C} \]

### Step 4: Calculate the effective current
The current efficiency is given as 90%. Therefore, the effective current (I) used in the calculation is:

\[ I = 9650 \, \text{A} \times 0.90 = 8685 \, \text{A} \]

### Step 5: Calculate the time required for electrolysis
Using the relationship between charge, current, and time (Q = I × t), we can rearrange to find time (t):

\[ t = \frac{Q}{I} \]

Substituting the values we found:

\[ t = \frac{1286666.67 \, \text{C}}{8685 \, \text{A}} \approx 148.14 \, \text{s} \]

### Conclusion
The time required to produce enough aluminum for a case of 24 cans is approximately **148.14 seconds**.

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