Mole Concept and Stoichiometry — Question 31

(a) The mass of 22.4 L of a gas at S.T.P. is equal to it's gram molecular mass.

308 cm³ of chlorine weighs 0.979 g

∴ 22,400 cm³ of chlorine will weigh

= $\dfrac{0.979}{308}$ × 22400 = 71.2 g

(b) Molar mass of H₂ = 2H = 2 x 1 = 2 g

2g H₂ at 1 atm has volume = 22.4 dm³

∴ 4 g H₂ at 1 atm will have volume 2 x 22.4 = 44.8 dm³

Now, For 4 g H₂

P₁ = 1 atm, V₁ = 44.8 dm³

P₂ = 4 atm, V₂ = ?

Using formula P₁V₁ = P₂V₂

$\text{V}_2 = \dfrac{\text{P}_1\text{V}_1}{\text{P}_2} \\[1em] \text{V}_2 = \dfrac{1 \times 44.8}{4} \\[1em] = \bold{11.2} \space \bold{dm^3}$

(c) Molar mass of oxygen in carbon dioxide = 2O = 2 x 16 = 32 g

Mass of oxygen in 22.4 litres of CO₂ = 32 g

∴ Mass of oxygen in 2.2 litres of CO₂

= $\dfrac{32}{22.4}$ x 2.2 = 3.14 g