Solved Sample Paper 1 — Question 14

(a) No. of moles of oxygen atoms = $\dfrac{12.044 \times 10^{23}}{6.022 \times 10^{23}}$ = 2 moles

Mass of oxygen atoms = no. of moles x atomic mass

= 2 x 16 = 32 g.

∴ The mass of 12.044 x 10²³ oxygen atoms is 32 grams.

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

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

Hence, vapour density is 14