The number of atoms of Na, H, S and 0 in 0.5 mole of sodium bisulphite are respectively.
(Given : Atomic mass of Na = 23 u, H = 1 u, S = 32 u and O = 16 u)

Answer

To solve the problem of finding the number of atoms of sodium (Na), hydrogen (H), sulfur (S), and oxygen (O) in 0.5 moles of sodium bisulfite (NaHSO₃), we can follow these steps:

### Step 1: Identify the chemical formula
The chemical formula for sodium bisulfite is NaHSO₃. This indicates that each molecule contains:
- 1 atom of Sodium (Na)
- 1 atom of Hydrogen (H)
- 1 atom of Sulfur (S)
- 3 atoms of Oxygen (O)

### Step 2: Determine the number of atoms in 0.5 moles
We know that 1 mole of any substance contains Avogadro's number of entities, which is approximately \(6.022 \times 10^{23}\).

### Step 3: Calculate the number of atoms for each element
- **Sodium (Na)**:
- In 1 mole of NaHSO₃, there is 1 atom of Na.
- Therefore, in 0.5 moles:
\[ \text{Number of Na atoms} = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23} \]

- **Hydrogen (H)**:
- In 1 mole of NaHSO₃, there is 1 atom of H.
- Therefore, in 0.5 moles:
\[ \text{Number of H atoms} = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23} \]

- **Sulfur (S)**:
- In 1 mole of NaHSO₃, there is 1 atom of S.
- Therefore, in 0.5 moles:
\[ \text{Number of S atoms} = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23} \]

- **Oxygen (O)**:
- In 1 mole of NaHSO₃, there are 3 atoms of O.
- Therefore, in 0.5 moles:
\[ \text{Number of O atoms} = 0.5 \times 3 \times 6.022 \times 10^{23} = 9.033 \times 10^{23} \]

### Step 4: Summarize the results
The number of atoms in 0.5 moles of sodium bisulfite (NaHSO₃) is:
- Sodium (Na): \(3.011 \times 10^{23}\)
- Hydrogen (H): \(3.011 \times 10^{23}\)
- Sulfur (S): \(3.011 \times 10^{23}\)
- Oxygen (O): \(9.033 \times 10^{23}\)

### Final Answer
The number of atoms of Na, H, S, and O in 0.5 moles of sodium bisulfite are respectively:
- Na: \(3.011 \times 10^{23}\)
- H: \(3.011 \times 10^{23}\)
- S: \(3.011 \times 10^{23}\)
- O: \(9.033 \times 10^{23}\)