Calculate number of electron present in `9.5" g of " PO_(4)^(3-)`?

Options

A. 6
B. ` 5 N_(A)`
C. `0.1 N_(A)`
D. `4.7 N_(A)`

Answer

To calculate the number of electrons present in 9.5 g of \( PO_4^{3-} \), we can follow these steps:

### Step 1: Calculate the number of moles of \( PO_4^{3-} \)
The number of moles is calculated using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Given:
- Mass of \( PO_4^{3-} = 9.5 \, g \)
- Molar mass of \( PO_4^{3-} = 95 \, g/mol \)

Now, substituting the values:
\[ \text{Number of moles} = \frac{9.5 \, g}{95 \, g/mol} = 0.1 \, \text{moles} \]

### Step 2: Calculate the number of ions in 0.1 moles of \( PO_4^{3-} \)
Using Avogadro's number, which is \( 6.022 \times 10^{23} \) ions/mol, we can find the total number of ions:
\[ \text{Number of ions} = \text{Number of moles} \times \text{Avogadro's number} \]
Substituting the values:
\[ \text{Number of ions} = 0.1 \, \text{moles} \times 6.022 \times 10^{23} \, \text{ions/mol} = 6.022 \times 10^{22} \, \text{ions} \]

### Step 3: Calculate the number of electrons in one \( PO_4^{3-} \) ion
To find the total number of electrons in one \( PO_4^{3-} \) ion, we consider:
- Phosphorus (P) has 15 electrons.
- Each oxygen (O) has 8 electrons, and there are 4 oxygen atoms.

Calculating the total number of electrons:
\[ \text{Total electrons} = \text{Electrons from P} + \text{Electrons from O} + \text{Charge} \]
\[ \text{Total electrons} = 15 + (4 \times 8) + 3 = 15 + 32 + 3 = 50 \, \text{electrons} \]

### Step 4: Calculate the total number of electrons in 9.5 g of \( PO_4^{3-} \)
Now, we can find the total number of electrons by multiplying the number of ions by the number of electrons per ion:
\[ \text{Total electrons} = \text{Number of ions} \times \text{Electrons per ion} \]
Substituting the values:
\[ \text{Total electrons} = 6.022 \times 10^{22} \, \text{ions} \times 50 \, \text{electrons/ion} = 3.011 \times 10^{24} \, \text{electrons} \]

### Final Answer
The total number of electrons present in 9.5 g of \( PO_4^{3-} \) is \( 3.011 \times 10^{24} \) electrons.

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