Mole Concept & Stoichiometry Miscellaneous Exercises — Question 46

Gram molecular mass of C₂H₄
= 2(12) + 4(1)
= 24 + 4 = 28 g

As,

28 g of C₂H₄ = 1 mole
∴ 1.4 g of C₂H₄ = $\dfrac{1}{28}$ x 1.4 = 0.05 moles

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

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

Vol. occupied by 1 mole = 22.4 lit
∴ Vol. occupied by 0.05 moles = 22.4 x 0.05 = 1.12 lit.

Hence, vol. occupied is 1.12 lit

(b)
$\text{Vapour density} = \dfrac{\text{Molecular weight}}{2} = \dfrac{28}{2} = 14$

Hence, vapour density is 14