Question

In Dumas method one gram of carbon compound gives $50 \,mL$ of $N _{2}$ at $300\, K$ and $740 \,mm\, Hg$ pressure. If the aqueous tension at $300\, K$ is $15\, mm\, Hg$, what is the percentage of nitrogen in it?

• A. 42
• B. 10.84
• C. 21.68
• D. 2.71

Answer 1

Answer: (A)

42

Answer 2

Given,

$p=740 \,mm$ of Hg dry gas pressure

$p_{1}=740-15=725\, mm \,Hg$

$V_{1}=50 \,mL$

$T_{1}=300 \,K , p_{2}=760 \,mm \,Hg , T_{2}=273 \,K$

From the combine gas equation,

$\frac{p_{1} V_{1}}{T_{1}}=\frac{p_{2} V_{2}}{T_{2}}$

$\Rightarrow \frac{725 \times 50}{300}=\frac{760 \times V_{2}}{273}$

$V_{2}=\frac{725 \times 50 \times 273}{300 \times 760}=43.4 \,mL$

$22.4 \,L$ of $N _{2} \longrightarrow 28 \,g$ of $N _{2}$ at STP $(22400 \,mL)$

$4.3 .4\, L$ of $N _{2} \to \frac{28 \times 43.4} {22400} =5.4 \times 10^{-2}\,g$ of $N_2$

$\%$ mass of $N _{2} =\frac{\text { Mass of } N _{2}}{\text { Mass of substance }} \times 100$ $=\frac{5.4 \times 10^{-2}}{ 1 } \times 100$

$=5.4 \%$