If 32 g of `O_(2)` contain `6 022 xx 10^23` molecules at NTP then 32 g of S, under conditions, will contain,

Options

A. `6.022 xx 10^(23)` S
B. `3.011 xx 10^(23)` S
C. `12.044 xx 10^(23) S`
D. `1 xx 10^(23) S`

Answer

To solve the problem, we need to determine how many molecules are present in 32 g of sulfur (S) under the same conditions as 32 g of oxygen (O₂) at Normal Temperature and Pressure (NTP).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We know that 32 g of O₂ contains \(6.022 \times 10^{23}\) molecules.
- This is based on the molar mass of O₂, which is approximately 32 g/mol.

2. **Finding the Molar Mass of Sulfur**:
- The molar mass of sulfur (S) is approximately 32 g/mol.
- This means that 32 g of sulfur is equivalent to 1 mole of sulfur.

3. **Using Avogadro's Number**:
- Avogadro's number (\(6.022 \times 10^{23}\)) tells us the number of molecules in one mole of any substance.
- Since we have 1 mole of sulfur (32 g), it will also contain \(6.022 \times 10^{23}\) molecules.

4. **Conclusion**:
- Therefore, 32 g of sulfur will contain \(6.022 \times 10^{23}\) molecules, just like 32 g of O₂.

### Final Answer:
32 g of sulfur will contain \(6.022 \times 10^{23}\) molecules.

---