Calculate no. of each atom present in `106.5 g` of`NaCIO_(3^(.)`

Explanation

To calculate the number of each atom present in 106.5 g of NaClO₃ (sodium chlorate), we can follow these steps:

### Step 1: Determine the Molar Mass of NaClO₃
The molar mass of sodium chlorate (NaClO₃) can be calculated by adding the atomic masses of its constituent elements:
- Sodium (Na): 23 g/mol
- Chlorine (Cl): 35.5 g/mol
- Oxygen (O): 16 g/mol (and there are 3 oxygen atoms)

Calculating the molar mass:
\[ \text{Molar mass of NaClO}_3 = 23 + 35.5 + (3 \times 16) = 23 + 35.5 + 48 = 106.5 \text{ g/mol} \]

### Step 2: Calculate the Number of Moles of NaClO₃
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
Substituting the values:
\[ \text{Number of moles of NaClO}_3 = \frac{106.5 \text{ g}}{106.5 \text{ g/mol}} = 1 \text{ mol} \]

### Step 3: Determine the Number of Atoms of Each Element
From the formula NaClO₃, we can see that:
- There is 1 sodium (Na) atom
- There is 1 chlorine (Cl) atom
- There are 3 oxygen (O) atoms

Using Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol), we can calculate the number of atoms for each element:
- Number of Na atoms:
\[ 1 \text{ mol NaClO}_3 \times 1 \text{ mol Na} \times 6.022 \times 10^{23} \text{ atoms/mol} = 6.022 \times 10^{23} \text{ Na atoms} \]
- Number of Cl atoms:
\[ 1 \text{ mol NaClO}_3 \times 1 \text{ mol Cl} \times 6.022 \times 10^{23} \text{ atoms/mol} = 6.022 \times 10^{23} \text{ Cl atoms} \]
- Number of O atoms:
\[ 1 \text{ mol NaClO}_3 \times 3 \text{ mol O} \times 6.022 \times 10^{23} \text{ atoms/mol} = 3 \times 6.022 \times 10^{23} = 1.8066 \times 10^{24} \text{ O atoms} \]

### Final Answer:
- Number of Na atoms: \(6.022 \times 10^{23}\)
- Number of Cl atoms: \(6.022 \times 10^{23}\)
- Number of O atoms: \(1.8066 \times 10^{24}\)