Time required to deposit one millimole of aluminium metal by the passage of 9.65 amp through aqueous solution of aluminium ion is:

Options

A. 30s
B. 10s
C. 30,000 s
D. 10,000 s

Explanation

To solve the problem of finding the time required to deposit one millimole of aluminum metal by the passage of 9.65 ampere through an aqueous solution of aluminum ions, we can follow these steps:

### Step 1: Understand the Reaction
The reaction for the deposition of aluminum from aluminum ions is:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]
This indicates that 1 mole of aluminum requires 3 moles of electrons.

### Step 2: Calculate the Charge Required for 1 Millimole of Aluminum
We know that 1 mole of electrons corresponds to 96500 coulombs (Faraday's constant). Therefore, for 1 mole of aluminum, the charge required is:
\[ Q = 3 \times 96500 \, \text{C} \]

For 1 millimole (which is \(10^{-3}\) moles):
\[ Q = 3 \times 96500 \times 10^{-3} \, \text{C} \]
\[ Q = 3 \times 96.5 \, \text{C} \]
\[ Q = 289.5 \, \text{C} \]

### Step 3: Use the Formula to Find Time
The relationship between charge (Q), current (I), and time (T) is given by:
\[ Q = I \times T \]
Rearranging this gives:
\[ T = \frac{Q}{I} \]

Substituting the values we have:
\[ T = \frac{289.5 \, \text{C}}{9.65 \, \text{A}} \]

### Step 4: Calculate the Time
Now, performing the calculation:
\[ T = \frac{289.5}{9.65} \approx 30.0 \, \text{seconds} \]

### Conclusion
The time required to deposit one millimole of aluminum metal by the passage of 9.65 ampere through an aqueous solution of aluminum ion is approximately **30 seconds**.

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