A current of 10 amp is passed through molten `AlCl_(3)` for 96.5 seconds. Calculate the mass of Al deposited.

Answer

To solve the problem of calculating the mass of aluminum deposited when a current of 10 A is passed through molten AlCl₃ for 96.5 seconds, we can follow these steps:

### Step 1: Calculate the total charge (Q) passed through the electrolyte.
The total charge can be calculated using the formula:
\[ Q = I \times t \]
where:
- \( I \) = current in amperes (A)
- \( t \) = time in seconds (s)

Given:
- \( I = 10 \, \text{A} \)
- \( t = 96.5 \, \text{s} \)

Calculating:
\[ Q = 10 \, \text{A} \times 96.5 \, \text{s} = 965 \, \text{C} \]

### Step 2: Determine the number of moles of electrons (n) transferred.
Using Faraday's law of electrolysis, the number of moles of electrons can be calculated using the formula:
\[ n = \frac{Q}{F} \]
where:
- \( F \) = Faraday's constant (approximately \( 96500 \, \text{C/mol} \))

Calculating:
\[ n = \frac{965 \, \text{C}}{96500 \, \text{C/mol}} = 0.01 \, \text{mol} \]

### Step 3: Relate moles of aluminum deposited to moles of electrons.
From the electrolysis of AlCl₃, we know that:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]
This means that 3 moles of electrons are required to deposit 1 mole of aluminum.

Thus, the moles of aluminum (Al) deposited can be calculated as:
\[ \text{Moles of Al} = \frac{n}{3} = \frac{0.01 \, \text{mol}}{3} = 0.00333 \, \text{mol} \]

### Step 4: Calculate the mass of aluminum deposited.
The mass of aluminum can be calculated using the formula:
\[ \text{mass} = \text{moles} \times \text{molar mass} \]
where the molar mass of aluminum (Al) is approximately 27 g/mol.

Calculating:
\[ \text{mass} = 0.00333 \, \text{mol} \times 27 \, \text{g/mol} = 0.08991 \, \text{g} \]

### Step 5: Round the mass to a suitable number of significant figures.
Rounding to three significant figures, we get:
\[ \text{mass} \approx 0.090 \, \text{g} \]

### Final Answer:
The mass of aluminum deposited is approximately **0.090 grams**.

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