The number of electrons required to deposit 1g equivalent aluminium (At. Wt. =27) from a solution of aluminium chloride will be

Options

A. 1
B. 2
C. 3
D. 4

Answer

To determine the number of electrons required to deposit 1 gram equivalent of aluminum from a solution of aluminum chloride, we can follow these steps:

### Step 1: Understand the concept of equivalent weight
The equivalent weight of an element is the mass of that element that combines with or displaces 1 mole of hydrogen or 1 mole of electrons.

### Step 2: Calculate the equivalent weight of aluminum
The atomic weight of aluminum (Al) is given as 27 g/mol. Aluminum has a valency of +3, which means it can lose 3 electrons to form Al³⁺ ions.

The equivalent weight of aluminum can be calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Atomic Weight}}{\text{Valency}} \]
Substituting the values:
\[ \text{Equivalent Weight} = \frac{27 \text{ g/mol}}{3} = 9 \text{ g/equiv} \]

### Step 3: Determine the number of equivalents in 1 g equivalent of aluminum
Since we are interested in 1 gram equivalent of aluminum, we can directly use the equivalent weight calculated above.

### Step 4: Calculate the number of moles of electrons required
To deposit 1 equivalent of aluminum, we need to consider how many electrons are required to reduce Al³⁺ to Al. Since aluminum has a valency of +3, it requires 3 electrons to be deposited:
\[ \text{Electrons required} = 3 \text{ electrons} \]

### Step 5: Conclusion
Thus, the number of electrons required to deposit 1 gram equivalent of aluminum from a solution of aluminum chloride is **3 electrons**.

### Final Answer
The number of electrons required to deposit 1 gram equivalent of aluminum is **3**.

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