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 problem step by step, we need to analyze the assertion (A) and the reason (R) provided in the question.

### Step 1: Understanding the Assertion
The assertion states that "Molten aluminium chloride when electrolyzed using 0.1F, deposits 0.1g equivalent of aluminium."

### Step 2: Understanding the Reason
The reason states that "Mass of substance deposited is directly proportional to the quantity of electricity." This is a fundamental principle of electrochemistry, known as Faraday's laws of electrolysis.

### Step 3: Applying Faraday’s Laws
According to Faraday's first law of electrolysis:
\[ W = Z \times Q \]
Where:
- \( W \) = mass of the substance deposited (in grams)
- \( Z \) = electrochemical equivalent (in grams per coulomb)
- \( Q \) = quantity of electricity (in coulombs)

### Step 4: Finding the Electrochemical Equivalent (Z)
The electrochemical equivalent \( Z \) can be calculated using the formula:
\[ Z = \frac{M}{nF} \]
Where:
- \( M \) = molar mass of aluminum = 27 g/mol
- \( n \) = number of electrons transferred (n-factor) = 3 for Al³⁺ to Al
- \( F \) = Faraday's constant = 96500 C/mol

Substituting the values:
\[ Z = \frac{27}{3 \times 96500} = \frac{27}{289500} \approx 9.33 \times 10^{-5} \text{ g/C} \]

### Step 5: Calculating the Quantity of Electricity (Q)
Given that the current is 0.1 Faraday, we convert this to coulombs:
\[ Q = 0.1 \times 96500 = 9650 \text{ C} \]

### Step 6: Calculating the Mass Deposited (W)
Now, we can calculate the mass of aluminum deposited:
\[ W = Z \times Q = (9.33 \times 10^{-5}) \times 9650 \approx 0.9 \text{ g} \]

### Step 7: Finding the Gram Equivalent
To find the gram equivalent of aluminum deposited:
\[ \text{Gram Equivalent} = \frac{W}{n} = \frac{0.9}{3} = 0.3 \text{ g equivalent} \]

### Step 8: Conclusion
The assertion states that 0.1 g equivalent of aluminum is deposited, but we calculated that 0.3 g equivalent is deposited. Therefore, the assertion is false. Since the reason is a true statement, but it does not support the assertion, the reason is also considered false in this context.

### Final Answer
Both the assertion (A) and the reason (R) are false.

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