Mole Concept and Stoichiometry — Question 33

Given:

P = 1140 mm Hg

Density = D = 3 g per L

T = 273 °C = 273 + 273 = 546 K

gram molecular mass = ?

At S.T.P., the volume of one mole of any gas is 22.4 L

Volume of unknown gas at S.T.P. = ?

By Charle’s law.

V₁ = 1 L

T₁ = 546 K

T₂ = 273 K

V₂ = ?

$\dfrac{\text{V}_1}{\text{T}_1}$ = $\dfrac{\text{V}_2}{\text{T}_2}$

Hence, V₂ = $\dfrac{1}{546}$ x 273 = 0.5 L

Volume at standard pressure = ?

Apply Boyle’s law.

P₁ = 1140 mm Hg

V₁ = 0.5 L

P₂ = 760 mm Hg

V₂ = ?

P₁ × V₁ = P₂ × V₂

V₂ = $\dfrac{1140 \times 0.5}{760}$ = 0.75 L

Now,

22.4 L = 1 mole of any gas at S.T.P.,

then 0.75 L = $\dfrac{0.75}{22.4}$

= 0.0335 moles

The original mass is 3 g

Molecular mass = $\dfrac{\text{Mass of compound}}{\text{Moles of compound}}$

= $\dfrac{3}{0.0335 }$ = 89.55 ≈ 89.6 g per mole

Hence, the gram molecular mass of the unknown gas is 89.6g