When 0.1 Faraday of electricity is passed in aqueous solution of
`AlCl_(3)`. The amount of Al deposited on cathode is

Options

A. 27g
B. 9g
C. 0.27g
D. 0.9g

Explanation

To solve the problem of how much aluminum (Al) is deposited on the cathode when 0.1 Faraday of electricity is passed through an aqueous solution of AlCl₃, we can follow these steps:

### Step 1: Understand the Ionization of AlCl₃
Aluminum chloride (AlCl₃) dissociates in water to form aluminum ions (Al³⁺) and chloride ions (Cl⁻):
\[ \text{AlCl}_3 \rightarrow \text{Al}^{3+} + 3\text{Cl}^- \]

### Step 2: Identify the Reduction Reaction
At the cathode, aluminum ions gain electrons to form solid aluminum:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al (s)} \]
This indicates that 3 moles of electrons are required to deposit 1 mole of aluminum.

### Step 3: Determine the Number of Moles of Electrons
We know that 1 Faraday corresponds to 1 mole of electrons (approximately 96500 coulombs). Therefore, 0.1 Faraday corresponds to:
\[ 0.1 \text{ Faraday} = 0.1 \text{ moles of electrons} \]

### Step 4: Calculate the Amount of Aluminum Deposited
Since 3 moles of electrons are needed to deposit 1 mole of aluminum, we can find out how many moles of aluminum can be deposited with 0.1 moles of electrons:
\[ \text{Moles of Al deposited} = \frac{0.1 \text{ moles of electrons}}{3} = \frac{0.1}{3} \text{ moles of Al} \]

### Step 5: Convert Moles of Aluminum to Grams
The molar mass of aluminum (Al) is approximately 27 g/mol. Therefore, the mass of aluminum deposited is:
\[ \text{Mass of Al} = \text{Moles of Al} \times \text{Molar mass of Al} \]
\[ \text{Mass of Al} = \left(\frac{0.1}{3}\right) \text{ moles} \times 27 \text{ g/mol} \]
\[ \text{Mass of Al} = \frac{0.1 \times 27}{3} \]
\[ \text{Mass of Al} = \frac{2.7}{3} = 0.9 \text{ grams} \]

### Final Answer
The amount of aluminum deposited on the cathode is **0.9 grams**.

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