Question

A gas cylinder contains $24\times {10}^{24}$molecules of nitrogen gas. If Avogadro's number is $6\times {10}^{23}$ and the relative atomic mass of nitrogen is 14, calculate :

(i) Mass of nitrogen gas in the cylinder

(ii) Volume of nitrogen at S.T.P. in dm³

Answer 1

(i) Molecular mass of nitrogen ${\mathrm{N}}_2$= $$\begin{gathered} 2\times 14 \\ =28g \end{gathered}$$

$\because 6.023\times {10}^{23}$ molecules of nitrogen weigh = 28 g

$\therefore 24\times {10}^{24}$ molecules of nitrogen will weigh = $\frac{28}{6.023\times {10}^{23}}\times 24\times {10}^{24}$

= $$\begin{gathered} 28\times 40 \\ =1120g \end{gathered}$$$$\begin{gathered} 2\times 14 \\ =28g \end{gathered}$$

(ii)$\because 6.023\times {10}^{23}$ molecules of nitrogen at S.T.P. occupy = 22.4 L

$\therefore 24\times {10}^{24}$ Molecules of nitrogen ${\mathrm{N}}_2$at S.T.P. occupy = $$\begin{gathered} \frac{22.4\times 24\times {10}^{24}}{6\times {10}^{23}} \\ =22.4\times 40 \\ =896d{m}^3 \end{gathered}$$