The number of electrons in 3.1 mg `NO_(3^(-))` is:

Options

A. 32
B. `1.6 xx 10^(-3)`
C. `9.6 xx 10^(20)`
D. `9.6 xx 10^(23)`

Answer

To find the number of electrons in 3.1 mg of the nitrate ion (NO₃⁻), we can follow these steps:

### Step 1: Calculate the molar mass of NO₃⁻
The molar mass of NO₃⁻ can be calculated as follows:
- Nitrogen (N) has a molar mass of approximately 14 g/mol.
- Oxygen (O) has a molar mass of approximately 16 g/mol, and there are 3 oxygen atoms in NO₃⁻.

So, the molar mass of NO₃⁻ is:
\[ \text{Molar mass of NO₃⁻} = 14 + (3 \times 16) = 14 + 48 = 62 \text{ g/mol} \]

### Step 2: Convert the mass of NO₃⁻ from mg to g
Given mass of NO₃⁻ is 3.1 mg. To convert this to grams:
\[ 3.1 \text{ mg} = 3.1 \times 10^{-3} \text{ g} \]

### Step 3: Calculate the number of moles of NO₃⁻
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles of NO₃⁻} = \frac{3.1 \times 10^{-3} \text{ g}}{62 \text{ g/mol}} \approx 5.0 \times 10^{-5} \text{ moles} \]

### Step 4: Calculate the number of nitrate ions
Using Avogadro's number (approximately \(6.022 \times 10^{23}\) entities/mol):
\[ \text{Number of NO₃⁻ ions} = \text{Number of moles} \times \text{Avogadro's number} \]
\[ \text{Number of NO₃⁻ ions} = 5.0 \times 10^{-5} \text{ moles} \times 6.022 \times 10^{23} \text{ ions/mol} \approx 3.01 \times 10^{19} \text{ ions} \]

### Step 5: Calculate the number of electrons in NO₃⁻
Each NO₃⁻ ion contains a total of 32 electrons (7 from nitrogen and 24 from three oxygen atoms, plus one additional electron due to the negative charge):
\[ \text{Total electrons per NO₃⁻} = 32 \]

### Step 6: Calculate the total number of electrons
\[ \text{Total electrons} = \text{Number of NO₃⁻ ions} \times \text{Total electrons per NO₃⁻} \]
\[ \text{Total electrons} = 3.01 \times 10^{19} \text{ ions} \times 32 \text{ electrons/ion} \approx 9.63 \times 10^{20} \text{ electrons} \]

### Final Answer
The total number of electrons in 3.1 mg of NO₃⁻ is approximately \(9.63 \times 10^{20}\).

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