Number of atoms of iron present in 100 gm `Fe_(2)O_(3)` having 20% purity is

Options

A. `0.2N_(A)`
B. `0.25N_(A)`
C. `0.5 N_(A)`
D. `0.3 N_(A)`

Answer

To find the number of atoms of iron present in 100 grams of `Fe_(2)O_(3)` with 20% purity, we can follow these steps:

### Step 1: Calculate the mass of pure `Fe_(2)O_(3)`
Since the sample has 20% purity, the mass of pure `Fe_(2)O_(3)` can be calculated as follows:

\[ \text{Mass of pure } Fe_2O_3 = \text{Total mass} \times \text{Purity} = 100 \, \text{g} \times \frac{20}{100} = 20 \, \text{g} \]

### Step 2: Calculate the molar mass of `Fe_(2)O_(3)`
The molar mass of `Fe_(2)O_(3)` can be calculated using the atomic masses of iron (Fe) and oxygen (O):

- Atomic mass of Fe = 55.85 g/mol
- Atomic mass of O = 16.00 g/mol

\[ \text{Molar mass of } Fe_2O_3 = (2 \times 55.85) + (3 \times 16.00) = 111.7 + 48.0 = 159.7 \, \text{g/mol} \approx 160 \, \text{g/mol} \]

### Step 3: Calculate the number of moles of `Fe_(2)O_(3)`
Using the mass of pure `Fe_(2)O_(3}` and its molar mass, we can find the number of moles:

\[ \text{Number of moles of } Fe_2O_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{20 \, \text{g}}{160 \, \text{g/mol}} = 0.125 \, \text{mol} \]

### Step 4: Calculate the number of moles of iron (Fe) in `Fe_(2)O_(3)`
From the formula `Fe_(2)O_(3)`, we see that there are 2 moles of iron for every mole of `Fe_(2)O_(3}`:

\[ \text{Number of moles of Fe} = 0.125 \, \text{mol} \times 2 = 0.25 \, \text{mol} \]

### Step 5: Calculate the number of atoms of iron
Using Avogadro's number (approximately \(6.022 \times 10^{23}\) atoms/mol), we can find the number of atoms of iron:

\[ \text{Number of atoms of Fe} = \text{Number of moles of Fe} \times \text{Avogadro's number} = 0.25 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \]

Calculating this gives:

\[ \text{Number of atoms of Fe} \approx 0.25 \times 6.022 \times 10^{23} \approx 1.505 \times 10^{23} \, \text{atoms} \]

### Final Answer
The number of atoms of iron present in 100 grams of `Fe_(2)O_(3}` with 20% purity is approximately \(1.505 \times 10^{23}\) atoms.

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