A: Molten aluminium chloride when electrolysed using 0.1F, deposits 0.1g equivalent of aluminium.
R: Mass of substance deposited `alpha` quantity of electricity.

Options

A. If both Assertion & Reason are true and the reason is the correct explanation of the assertion, then mark (1)
B. If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2)
C. If Assertion is true statement but Reason is false, then mark (3)
D. If both Assertion and Reason are false statements then mark (4)

Answer

To solve the question regarding the electrolysis of molten aluminum chloride and the deposition of aluminum, we will break down the steps involved in the calculation.

### Step-by-Step Solution:

1. **Understanding the Electrolysis Reaction:**
The electrolysis of molten aluminum chloride (AlCl₃) can be represented by the following half-reaction:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]
This indicates that 1 mole of aluminum ions (Al³⁺) requires 3 moles of electrons (3 Faradays) to deposit 1 mole of aluminum.

2. **Determine the Equivalent Weight (Z) of Aluminum:**
The equivalent weight (Z) can be calculated using the formula:
\[ Z = \frac{\text{Molar Mass}}{n} \]
where:
- Molar Mass of aluminum (Al) = 27 g/mol
- n (number of electrons transferred) = 3

Thus,
\[ Z = \frac{27}{3} = 9 \text{ g/equiv} \]

3. **Calculating the Quantity of Electricity (Q):**
Given that the electrolysis is performed using 0.1 Faraday (F), we can calculate the total charge (Q):
\[ Q = n \times F \]
where F (Faraday's constant) = 96500 C/mol. Therefore,
\[ Q = 0.1 \times 96500 = 9650 \text{ C} \]

4. **Calculating the Mass of Aluminum Deposited:**
The mass of aluminum deposited (m) can be calculated using the formula:
\[ m = Z \times Q \]
Substituting the values:
\[ m = 9 \text{ g/equiv} \times 0.1 \text{ equiv} = 0.9 \text{ g} \]

5. **Finding the Gram Equivalent:**
The gram equivalent of aluminum deposited can be calculated as follows:
\[ \text{Gram Equivalent} = \frac{\text{Mass Deposited}}{Z} \]
Substituting the values:
\[ \text{Gram Equivalent} = \frac{0.9 \text{ g}}{9 \text{ g/equiv}} = 0.1 \text{ equiv} \]

6. **Conclusion:**
The assertion states that 0.1 Faraday deposits 0.1 g equivalent of aluminum, which is incorrect since we calculated that it deposits 0.1 equivalent of aluminum but the mass is 0.9 g, not 0.1 g.

The reason states that the mass of substance deposited is directly proportional to the quantity of electricity, which is true. However, since the assertion is false, the overall conclusion is that both assertion and reason are incorrect.

### Final Answer:
Both the assertion and the reason are incorrect.

---