Number of atoms of oxygen present in 10.6 g of `Na_(2)CO_(3)` will be

Options

A. `6.02xx10^(22)`
B. `12.04xx10^(22)`
C. `1.806xx10^(23)`
D. `31.80xx10^(28)`

Explanation

To find the number of oxygen atoms present in 10.6 g of sodium carbonate (Na₂CO₃), we can follow these steps:

### Step 1: Calculate the molar mass of sodium carbonate (Na₂CO₃).
- Sodium (Na) has an atomic mass of 23 g/mol. Since there are 2 sodium atoms:
\[ 2 \times 23 = 46 \text{ g/mol} \]
- Carbon (C) has an atomic mass of 12 g/mol. There is 1 carbon atom:
\[ 1 \times 12 = 12 \text{ g/mol} \]
- Oxygen (O) has an atomic mass of 16 g/mol. There are 3 oxygen atoms:
\[ 3 \times 16 = 48 \text{ g/mol} \]
- Now, add these together to find the total molar mass of Na₂CO₃:
\[ 46 + 12 + 48 = 106 \text{ g/mol} \]

### Step 2: Calculate the number of moles of Na₂CO₃ in 10.6 g.
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles} = \frac{10.6 \text{ g}}{106 \text{ g/mol}} \approx 0.1 \text{ moles} \]

### Step 3: Determine the number of moles of oxygen in Na₂CO₃.
In one mole of Na₂CO₃, there are 3 moles of oxygen. Therefore, the number of moles of oxygen in 0.1 moles of Na₂CO₃ is:
\[ \text{Moles of O} = 0.1 \text{ moles of Na₂CO₃} \times 3 = 0.3 \text{ moles of O} \]

### Step 4: Convert moles of oxygen to atoms of oxygen.
Using Avogadro's number (6.022 × 10²³ atoms/mole):
\[ \text{Number of oxygen atoms} = 0.3 \text{ moles of O} \times 6.022 \times 10^{23} \text{ atoms/mole} \]
Calculating this gives:
\[ \text{Number of oxygen atoms} \approx 1.806 \times 10^{23} \text{ atoms} \]

### Final Answer:
The number of oxygen atoms present in 10.6 g of sodium carbonate (Na₂CO₃) is approximately \(1.806 \times 10^{23}\) atoms.

---