In Dumas method one gram of carbon compound gives \(50 \mathrm{~mL}\) of \(\mathrm{N}_2\) at \(300…
In Dumas method one gram of carbon compound gives \(50 \mathrm{~mL}\) of \(\mathrm{N}_2\) at \(300 \mathrm{~K}\) and \(740 \mathrm{~mm} \mathrm{Hg}\) pressure. If the aqueous tension at \(300 \mathrm{~K}\) is \(15 \mathrm{~mm} \mathrm{Hg}\), what is the percentage of nitrogen in it?
5
10.84
21.68
2.71
Solution
Given,
\(p=740 \mathrm{~mm}\) of \(\mathrm{Hg}\) dry gas pressure
\(p_1=740-15=725 \mathrm{~mm} \mathrm{Hg}\)
\(V_1=50 \mathrm{~mL}\)
\(T_1=300 \mathrm{~K}, p_2=760 \mathrm{~mm} \mathrm{Hg}, T_2=273 \mathrm{~K}\)
From the combine gas equation,
\(\begin{gathered}
\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2} \\
\Rightarrow \frac{725 \times 50}{300}=\frac{760 \times V_2}{273}
\end{gathered}\)
\(V_2=\frac{725 \times 50 \times 273}{300 \times 760}=43.4 \mathrm{~mL}\)
\(22.4 \mathrm{~L}\) of \(\mathrm{N}_2 \longrightarrow 28 \mathrm{~g}\) of \(\mathrm{N}_2\) at STP (22400 mL)
\(43.4 \mathrm{~L}_{\text {of } \mathrm{N}_2} \longrightarrow \frac{28 \times 434}{22400}=5.4 \times 10^{-2} \mathrm{~g}\) of \(\mathrm{N}_2\)
\(\%\) mass of \(\mathrm{N}_2=\frac{\text { Mass of } \mathrm{N}_2}{\text { Mass of substance }} \times 100\)
\(=\frac{5.4 \times 10^{-2}}{1} \times 100=5.4 \%\)
Hence, option (1) is correct.