The amount of electricity required to deposit one gram atom of Mg from its molten salt is

Options

A. 96500 C
B. `1.93 xx 10^(5)`C
C. 1 C
D. 4825 C

Answer

To determine the amount of electricity required to deposit one gram atom of magnesium (Mg) from its molten salt, we can follow these steps:

### Step 1: Understand the Reduction Process
When magnesium is deposited from its molten state, it undergoes a reduction reaction. In this process, magnesium ions (Mg²⁺) gain two electrons (2e⁻) to form solid magnesium (Mg). The half-reaction can be represented as:
\[ \text{Mg}^{2+} + 2e^- \rightarrow \text{Mg} \]

### Step 2: Determine the Moles of Magnesium
We know that 1 gram atom of magnesium corresponds to 1 mole of magnesium. The molar mass of magnesium (Mg) is approximately 24.31 g/mol. Therefore, 1 gram of magnesium is:
\[ \text{Moles of Mg} = \frac{1 \text{ g}}{24.31 \text{ g/mol}} \approx 0.041 \text{ mol} \]

### Step 3: Calculate the Number of Electrons Required
Since the reduction of one mole of magnesium ions requires 2 moles of electrons, for 1 mole of magnesium, the total moles of electrons required is:
\[ \text{Moles of electrons} = 2 \times 0.041 \text{ mol} \approx 0.082 \text{ mol} \]

### Step 4: Convert Moles of Electrons to Charge
Using Faraday's constant, which is approximately 96500 Coulombs per mole of electrons, we can calculate the total charge (Q) required:
\[ Q = \text{Moles of electrons} \times \text{Faraday's constant} \]
\[ Q = 0.082 \text{ mol} \times 96500 \text{ C/mol} \]

### Step 5: Perform the Calculation
Now, calculating the total charge:
\[ Q \approx 0.082 \times 96500 \approx 7919 \text{ C} \]

### Step 6: Convert to Scientific Notation
To express this in scientific notation:
\[ Q \approx 7.92 \times 10^3 \text{ C} \]

### Final Answer
The amount of electricity required to deposit one gram atom of magnesium from its molten salt is approximately:
\[ 7.92 \times 10^3 \text{ Coulombs} \]

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