Question

(a) Calculate the number of moles and number of molecules present in 1.4 g of ethylene gas. What is the volume occupied by the same amount of ethylene?
(b) What is the vapour density of ethylene?
(Avogadro's number = 6 × 10²³; Atomic weight of C = 12, H = 1; Molar volume = 22.4 litres at STP)

Answer 1

(a) Number of moles of ethylene in 1.4 g = $\frac{\mathrm{Mass}\mathrm{of}\mathrm{ethylene}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{ethylene}}=\frac{1.4}{28}=0.05$
Number of molecules of ethylene in 0.05 moles = Number of moles $\times$ Avogadro's Number
= 0.05 $\times$ 6 $\times$ 10²³ = 3 $\times$ 10²²
Volume occupied by 0.05 moles of ethylene = (0.05 $\times$ 22.4) L = 1.12 L

(b) Molecular mass of ethylene = 2 $\times$ Vapour density of ethylene
28 = 2 $\times$ Vapour density
Vapour density = 14