96500 coulombs deposit 107.9 g of Ag from its soluton. If `e = 1.6 xx 10^(-19)` coulomb, calculate the number of electrons per mole of electrons

Options

A. 96500
B. `1.6 xx 10^(-19)`
C. `6.02 xx 10^(23)`
D. `6.02 xx 10^(-23)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between charge, moles of electrons, and Faraday's constant.
Faraday's constant (F) is defined as the amount of electric charge per mole of electrons. The value of Faraday's constant is approximately 96500 coulombs per mole of electrons.

### Step 2: Calculate the number of moles of silver deposited.
The molar mass of silver (Ag) is approximately 107.9 g/mol. We can calculate the number of moles of silver deposited using the formula:

\[ \text{Number of moles of Ag} = \frac{\text{mass of Ag}}{\text{molar mass of Ag}} = \frac{107.9 \, \text{g}}{107.9 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 3: Relate the number of moles of silver to the number of moles of electrons.
In the electrolysis of silver nitrate (AgNO₃), each mole of silver deposited corresponds to one mole of electrons. Therefore, if 1 mole of silver is deposited, then 1 mole of electrons is also involved.

### Step 4: Calculate the total charge required to deposit 1 mole of silver.
Since we know that 1 mole of electrons corresponds to a charge of 96500 coulombs, we can say:

\[ \text{Charge for 1 mole of electrons} = 96500 \, \text{C} \]

### Step 5: Calculate the number of electrons per mole of electrons.
We know that the charge of a single electron (e) is given as \(1.6 \times 10^{-19}\) coulombs. To find the number of electrons (n) in one mole of electrons, we use the formula:

\[ n = \frac{\text{Charge per mole}}{\text{Charge of one electron}} = \frac{96500 \, \text{C}}{1.6 \times 10^{-19} \, \text{C/electron}} \]

Calculating this gives:

\[ n = \frac{96500}{1.6 \times 10^{-19}} \approx 6.02 \times 10^{23} \, \text{electrons} \]

### Final Answer:
The number of electrons per mole of electrons is approximately \(6.02 \times 10^{23}\).

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