Calculate the volume occupied by 3.5 g of O₂ gas at 27 °C and 740 mm pressure. [O = 16]

Gram molecular mass of O₂ = 2 x 16 = 32 g

1 mole of O₂ weighs 32 g and occupies 22.4 lit. vol.

∴ 3.5 g of O₂ occupies = $\dfrac{22.4}{32} \times 3.5 = 2.45 \text{ lit.}$

Volume occupied by 3.5 g of O₂ gas at 27°C and 740 mm pressure:

Using the gas equation,

$\dfrac{P${1}V{1}}{T{1}} = \dfrac{P{2}V{2}}{T{2}}

Substituting the values we get,

$\dfrac{760 \times 2.45}{273} = \dfrac{740 \times x}{300} \\[0.5em] x = \dfrac{760 \times 2.45 \times 300}{740 \times 273 } \\[0.5em] x = \dfrac{5,58,600}{2,02,020} \\ \\[0.5em] x = 2.76 \text{ lit}$

Hence, the volume occupied by 3.5 g of O₂ gas at 27°C and 740 mm pressure is 2.76 lit.