Question

In Dumas' method for estimation of nitrogen, 0.3g of an organic compound gave 50mL of nitrogen collected at 300 K temperature and 715 mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300 $K=15$ mm)

• A. 17.5%
• B. 28%
• C. 6.25%
• D. 31%

Answer 1

Answer: (C)

6.25%

Here,
Mass of organic substance $=0.3$ $g$
Volume of Nitrogen collected $=50$ $mL$
Temperature $\left(T_1\right)=300$ $K$
Vapour pressure of water $=15$ $mm$
Actual pressure of dry nitrogen $=715-15=700$ $mm$
At STP, we know that $P_2=760$ $mm$ and $T_2=273$ $K$ then $V_2=?$
$⟹\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
$⟹\frac{700\times 50}{300}=\frac{760\times V_2}{273}$
$⟹V_2=41.9$ $mL$
Hence, $\%$ of nitrogen $=\frac{28\times 41.9\times 100}{22400}=5.238\%\simeq 6.25\%$