Chemistry - Nomenclature Question with Solution | TestHub

Volume of nitrogen collected at 300 K and 725 mm pressure is 40 mL actual pressure
$=725-25=700$$\mathrm{m}\mathrm{m}$
Using$\frac{\stackrel{\underset{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s}}{\mathrm{G}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}}}{{\mathrm{P}}_1{\mathrm{V}}_1}}{{\mathrm{T}}_1}=\frac{\stackrel{\mathrm{A}\mathrm{t}\mathrm{S}\mathrm{T}\mathrm{P}}{{\mathrm{P}}_2{\mathrm{V}}_2}}{{\mathrm{T}}_2}$
${\mathrm{V}}_2=\frac{{\mathrm{P}}_1{\mathrm{V}}_1\times {\mathrm{T}}_2}{{\mathrm{T}}_1\times {\mathrm{P}}_2}$
Volume of nitrogen at STP$\left({\mathrm{V}}_2\right)=\frac{273\times 700\times 40}{300\times 760}=33.52\mathrm{m}\mathrm{L}$
As 22,400 mL of${\mathrm{N}}_2$at STP weight$=28\mathrm{g}$
So, 33.5 mL of nitrogen weight$=\frac{28\times 33.52}{22400}\mathrm{g}$
N%$=\frac{28\times {\mathrm{V}}_2\times 100}{22400\times \mathrm{W}}$
Percentage of nitrogen (N%)$=\frac{28\times 33.52\times 100}{22400\times 0.25}=16.76\mathrm{\%}$