At STP, the number of molecules present in 5.6 L of O₂ gas is:

Correct option is B

The correct answer (b) 1.5055 × 10²³

• At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 L and contains 6.022 × 10²³ molecules (Avogadro’s number).

• Given volume = 5.6 L

Now,  

​$$\text{Number of moles} = \frac{5.6}{22.4} = 0.25\ \text{mol} \\[10pt]\text{Number of molecules} = 0.25 \times 6.022 \times 10^{23} = 1.5055 \times 10^{23}$$

• The calculation shows that 5.6 L of O₂ gas at STP corresponds to 0.25 mol.

• Multiplying by Avogadro’s number gives: $$0.25 \times 6.022 \times 10^{23} = 1.5055 \times 10^{23}$$

Information Booster:

• STP refers to 0°C (273.15 K) and 1 atm pressure.

• 1 mole = 22.4 L at STP for ideal gases.

• Avogadro’s number: 6.022 × 10²³ molecules/mol

• Volume-to-moles conversions are key in gas stoichiometry.