The number of coulombs required for the deposition of 107.870 of silver is

Options

A. 96500
B. 48250
C. 193000
D. 10000

Explanation

To find the number of coulombs required for the deposition of 107.870 grams of silver (Ag), we can follow these steps:

### Step 1: Determine the molar mass of silver
The molar mass (or atomic weight) of silver (Ag) is approximately 107.87 g/mol.

### Step 2: Calculate the number of moles of silver
To find the number of moles of silver in 107.870 grams, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles of Ag} = \frac{107.870 \text{ g}}{107.87 \text{ g/mol}} = 1 \text{ mol} \]

### Step 3: Determine the charge required for deposition
The deposition of silver from its ion (Ag⁺) to solid silver (Ag) involves the transfer of one electron per atom of silver. Therefore, for 1 mole of silver, we need 1 mole of electrons.

### Step 4: Calculate the total charge in coulombs
According to Faraday's law of electrolysis, 1 mole of electrons corresponds to 1 Faraday of charge, which is approximately 96500 coulombs. Thus, for 1 mole of silver:
\[ \text{Charge required} = 1 \text{ mol} \times 96500 \text{ C/mol} = 96500 \text{ C} \]

### Conclusion
The number of coulombs required for the deposition of 107.870 grams of silver is **96500 coulombs**.

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