# Question ad42b

Volume of oxygen at STP present in 2L air is =2Lxx21%=0.42L#
Number of moles of ${O}_{2}$ present in $0.42 L$ oxygen at STP will be
$= \text{volume of oxygen"/"molar volume of the gas}$
$= \frac{0.42 L}{22.4 \frac{L}{\text{mol}}} = 18.75 \times {10}^{-} 3 m o l$