Calculate the no. of moles and the no. of molecules present in 1.4 g of ethylene gas (C2H4). What is the vol. occupied by the same amount of C2H4. State the vapour density of C2H4.

[Avog. No. = 6 × 10^23 ; C = 12, H = 1]

Gram molecular mass of C₂H₄

= 2(12) + 4(1)

= 24 + 4 = 28 g

28 g of C₂H₄ = 1 mole

∴ 1.4 g of C₂H₄ = 1\over28 × 1.4 = 0.05 moles

1 mole = 6 × 10²³ molecules

∴ 0.05 moles = 6 × 10²³ x 0.05 = 3 × 10²² molecules

Hence, no. of moles is 0.05 and no. of molecules is 3 × 10²².

Vol. occupied by 1 mole = 22.4 lit

∴ Vol. occupied by 0.05 moles = 22.4 × 0.05 = 1.12 lit.

Vapour density = molecular\ weight = 28\ 2 = 14

Hence, vol. occupied is 1.12 lit and vapour density is 14