The cost of electricity required to deposit `1g Mg` is Rs. 5.00. the cost of `30g` of `Al` to be deposited is Rs. `X`. Find the value of `x/40`?

Explanation

To solve the problem, we need to find the cost of depositing 30 grams of aluminum (Al) based on the cost of depositing 1 gram of magnesium (Mg). Here’s the step-by-step solution:

### Step 1: Determine the n-factor of Magnesium (Mg)
- Magnesium (Mg) has a valency of +2, meaning it requires 2 electrons to deposit 1 mole of Mg.
- Therefore, the n-factor of Mg is 2.

### Step 2: Calculate the equivalent mass of Magnesium
- The molar mass of Mg is 24 g/mol.
- The equivalent mass of Mg is calculated as:
\[ \text{Equivalent mass of Mg} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{24 \text{ g}}{2} = 12 \text{ g} \]

### Step 3: Determine the charge required for depositing 1 gram of Magnesium
- Since 12 grams of Mg requires 1 Faraday (F) of charge, the charge required for depositing 1 gram of Mg is:
\[ \text{Charge for 1 g of Mg} = \frac{1 \text{ F}}{12} \text{ Faraday} \]

### Step 4: Establish the cost of depositing 1 gram of Magnesium
- The cost to deposit 1 gram of Mg is given as Rs. 5.00. Therefore, we can write:
\[ \text{Cost for } \frac{1}{12} \text{ F} = 5 \text{ Rs.} \]

### Step 5: Determine the n-factor of Aluminum (Al)
- Aluminum (Al) has a valency of +3, meaning it requires 3 electrons to deposit 1 mole of Al.
- Therefore, the n-factor of Al is 3.

### Step 6: Calculate the equivalent mass of Aluminum
- The molar mass of Al is 27 g/mol.
- The equivalent mass of Al is calculated as:
\[ \text{Equivalent mass of Al} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{27 \text{ g}}{3} = 9 \text{ g} \]

### Step 7: Determine the charge required for depositing 30 grams of Aluminum
- Since 9 grams of Al requires 1 Faraday (F) of charge, the charge required for depositing 30 grams of Al is:
\[ \text{Charge for 30 g of Al} = \frac{30 \text{ g}}{9 \text{ g}} \text{ F} = \frac{30}{9} \text{ F} = \frac{10}{3} \text{ F} \]

### Step 8: Establish the cost for depositing 30 grams of Aluminum
- Let the cost for depositing 30 grams of Al be Rs. X.
- We know that the cost for 1 Faraday is Rs. 5 (from the cost of Mg).
- Therefore, the cost for depositing \(\frac{10}{3}\) Faraday is:
\[ \text{Cost for } \frac{10}{3} \text{ F} = \frac{10}{3} \times 5 = \frac{50}{3} \text{ Rs.} \]
- Thus, we can equate this to X:
\[ X = \frac{50}{3} \text{ Rs.} \]

### Step 9: Calculate the value of \( \frac{X}{40} \)
- Now, we need to find \( \frac{X}{40} \):
\[ \frac{X}{40} = \frac{\frac{50}{3}}{40} = \frac{50}{120} = \frac{5}{12} \text{ Rs.} \]

### Final Answer
Thus, the value of \( \frac{X}{40} \) is \( \frac{5}{12} \) Rs.

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