Question

In Dumas method for estimation of nitrogen, $0.3$ g of an organic compound gave $24.63$ mL of nitrogen collected at $300$ K temperature and $775$ mm pressure. The percentage composition of nitrogen in the compound is (aqueous tension at $300$ K = $15$ mm) :

• A. $8.9\%$
• B. $8.5\%$
• C. $9.3\%$
• D. $9.0\%$

Answer 1

Answer: (C)

$9.3\%$

Given : $m=0.3g$, $P_0=760mm$ $T_0=273K$

$V_0=?$, $P=775mm$, $f=15mm$

Solution : $P_1=P-f=775-15=760mm$

$V_1=24.63mL$, $T_1=300K$

$\frac{P_0{V}_0}{T_0}=\frac{P_1{V}_1}{T_1}$

Hence, $V_0=\frac{P_1{V}_1}{T_1}\times \frac{T_0}{P_0}=\frac{760\times 24.63}{300}\times \frac{273}{760}=22.4ml$

22400 ml of nitrogen at STP weighs 28 g.

Hence, 22.4 ml of nitrogen at STP will weigh $\frac{28\times 22.4}{22400}g$ of nitrogen.

100 g of compound will contain $\frac{28\times 22.4}{22400}\times \frac{100}{0.3}=9.3$% of nitrogen.