(a) Calculate the mass of substance 'A' which in gaseous form occupies 10 litres at 27°C and 700 mm pressure. The molecular mass of 'A' is 60.

(b) A gas occupied 360 cm³ at 87°C and 380 mm Hg pressure. If the mass of gas is 0.546 g, find it's relative molecular mass.

(a) Given, molecular mass of 'A' is 60

V₁ = 10 L

T₁ = 27 + 273 K = 300 K

P₁ = 700 mm

T₂ = 273 K

P₂ = 760 mm

V₂ = ?

Using formula:

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

Substituting in the formula,

$\dfrac{700 \times 10}{300}$ = $\dfrac{760 \times\text{V}_2}{273}$

V₂ = $\dfrac{700 \times10 \times 273}{300 \times760}$ = $\dfrac{1911}{228}$ = 8.38 L

As, 22.4 L of A weighs 60 g

∴ 8.38 L of A will weigh $\dfrac{60}{22.4}$ x 8.38

= 22.446 = 22.45 g

(b) V₁ = 360 cm³

T₁ = 87 + 273 K = 360 K

P₁ = 380 mm Hg pressure

T₂ = 273 K

P₂ = 760 mm Hg pressure

V₂ = ?

Using formula:

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

Substituting in the formula,

$\dfrac{380\times 360}{360}$ = $\dfrac{760 \times\text{V}_2}{273}$

V₂ = $\dfrac{380 \times 273}{760}$ = $\dfrac{10,374}{760}$ = 136.5 cm³

136.5 cm³ of gas weigh = 0.546 g

22400 cm³ of gas weigh $\dfrac{0.546}{136.5}$ x 22400

= 89.6 amu