A coin is made up of Al and weighs 0.75 g. It has a square shape and its diagonal measures 17 mm. It is electrically neutral and contains equal amounts of positive and negative charges. The magnitude of these charges is (Atomic mass of Al = 26.98 g)

Options

A. `3.47xx10^(4)C`
B. `3.47xx10^(2)C`
C. `1.67xx10^(20)C`
D. `1.67xx10^(22)C`

Explanation

To find the magnitude of the positive and negative charges in the aluminum coin, we can follow these steps:

### Step 1: Calculate the number of moles of aluminum in the coin.
The mass of the aluminum coin is given as 0.75 g, and the molar mass of aluminum (Al) is 26.98 g/mol.

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.75 \, \text{g}}{26.98 \, \text{g/mol}} \approx 0.0278 \, \text{mol} \]

### Step 2: Calculate the number of atoms (or molecules) of aluminum in the coin.
Using Avogadro's number, which is \(6.022 \times 10^{23} \, \text{atoms/mol}\):

\[ \text{Number of atoms} = \text{Number of moles} \times \text{Avogadro's number} = 0.0278 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 1.67 \times 10^{22} \, \text{atoms} \]

### Step 3: Calculate the total number of protons (or electrons) in the aluminum atoms.
Each aluminum atom contains 13 protons and 13 electrons. Therefore, the total number of protons (or electrons) in the coin is:

\[ \text{Total protons} = 13 \times \text{Number of atoms} = 13 \times 1.67 \times 10^{22} \approx 2.17 \times 10^{23} \, \text{protons} \]

### Step 4: Calculate the total charge from the protons (or electrons).
The charge of one proton (or electron) is \(1.6 \times 10^{-19} \, \text{C}\). Therefore, the total charge is:

\[ \text{Total charge} = \text{Total protons} \times \text{Charge of one proton} = 2.17 \times 10^{23} \times 1.6 \times 10^{-19} \approx 3.47 \times 10^{4} \, \text{C} \]

### Final Answer:
The magnitude of the positive and negative charges in the aluminum coin is approximately \(3.47 \times 10^{4} \, \text{C}\).

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