Question

A gas cylinder contains 24 × 10²⁴ molecules of nitrogen gas. If Avogadro’s number is 6 × 10²³ and the relative atomic mass of nitrogen is 14, calculate:

1. Mass of nitrogen gas in the cylinder.
2. Volume of nitrogen at STP in dm³.

Answer 1

i. Nitrogen exists as N₂.

∴ Molecular mass of nitrogen = 14 × 2

= 28 g

6 × 10²³ molecules of nitrogen weigh = 28 g

∴ 24 × 10²⁴ molecule of nitrogen will weigh = `28/(6 xx 10^23) xx 24 xx 10^24`

= 28 × 4 × 10

= 1120 g

ii. As 6 × 10²³ molecules of nitrogen occupies = 22.4 dm³ at STP

1 molecule will occupy a volume of `22.4/(6 xx 10^23)` dm³.

∴ 24 × 10²⁴ molecules will occupy = `22.4/(6 xx 10^23) xx 24 xx 10^24`

= 895.9 dm³