Question

In Dumas method of estimation of nitrogen, $0.35\text{g}$ of an organic compound gave $55\text{mL}$ of nitrogen collected at $300\text{K}$ temperature and $715\text{mm}$ pressure. Find the percentage composition of nitrogen in the compound. $(\text{Aqueous tension at}300\text{K}=15\text{mm Hg})$

• A. $15.45$
• B. $16.45$
• C. $17.45$
• D. $14.45$

Answer 1

The correct option is B $16.45$

The parameters given in the question are as follows:
$\text{Weight of the organic compound taken}=0.35\text{g}$
Volume of $N_2$ collected at $300\text{K and}715\text{mm}$pressure $\left(V_1\right)=55\text{mL}$
$\text{Aqueous tension at}300\text{K}=15\text{mm Hg}$
$\text{Actual pressure,}{P}_1=715-15=700\text{mm Hg}$
Temperature at experimental conditions, $T_1=300K$
From the combined gas equation, we can write,
$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
Where, $P_2,V_2,\text{and}{T}_2$ represent pressure, volume, and temperature at STP conditions.
Volume of nitrogen at STP, $V_2=\frac{273\times 700\times 55}{300\times 760}=46.09\text{mL}$
We know that $22400\text{mL}$ at STP weighs $28\text{g}{N}_2$
$46.09\text{mL}\text{of}{N}_2\text{weighs}=\frac{28\times 46.09}{22400}\text{g}{N}_2$
The percentage of $N_2$ in $0.35\text{g}$ of the organic compound is given as:
Percentage of nitrogen $=\frac{28\times 46.09\times 100}{22400\times 0.35}=16.45\text{\%}$
Hence, option (b) is the correct answer.