Calculate the amount of electricity required to deposite 0.9 g of aluminium by electrolysis of a salt containing its ion, if the electrode reaction is
`Al^(3+)+3e^(-)rarrAl`,
(atomic mass of Al `=27, 1F=96500C`)

Options

A. `9.65xx10^(3)C`
B. `1.93xx10^(4)C`
C. `9.65xx10^(4)C`
D. `4.32 xx10^(5)C`

Answer

To calculate the amount of electricity required to deposit 0.9 g of aluminum by electrolysis, we can follow these steps:

### Step 1: Determine the number of moles of aluminum
First, we need to calculate the number of moles of aluminum (Al) that corresponds to 0.9 g.

\[ \text{Number of moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{0.9 \, \text{g}}{27 \, \text{g/mol}} = 0.03333 \, \text{mol} \]

### Step 2: Determine the number of electrons required
From the electrode reaction given:

\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]

We see that 3 moles of electrons are required to deposit 1 mole of aluminum. Therefore, for 0.03333 moles of aluminum, the number of moles of electrons required is:

\[ \text{Number of moles of electrons} = 0.03333 \, \text{mol Al} \times 3 \, \text{mol e}^- / \text{mol Al} = 0.1 \, \text{mol e}^- \]

### Step 3: Calculate the total charge in coulombs
Using Faraday's constant, which is the charge per mole of electrons (1 F = 96500 C), we can find the total charge required:

\[ \text{Total charge (Q)} = \text{Number of moles of electrons} \times \text{Faraday's constant} = 0.1 \, \text{mol} \times 96500 \, \text{C/mol} = 9650 \, \text{C} \]

### Step 4: Conclusion
The amount of electricity required to deposit 0.9 g of aluminum is:

\[ Q = 9650 \, \text{C} \]

### Final Answer
The answer is \( 9.65 \times 10^3 \, \text{C} \).

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