Question

In Duma's method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is:

• A. 16.76
• B. 15.76
• C. 17.36
• D. 18.2

Answer 1

The correct option is A 16.76

Given:

$m=0.25$ g, $V_1=40$ ml, $T_1=300$ K, $P_1=725mm-25mm=700mm$.

$P_0=760$ mm, $T_0=273$ K, $V_0=$ ?

$V_0=\frac{P_1{V}_1}{T_1}\times \frac{T_o}{P_0}=\frac{700mm\times 40ml}{300K}\times \frac{273K}{760mm}=33.53ml$.

At STP, 22400 ml of nitrogen occupies 22400 ml.

22400mL OF $N_2$ at STP weighs = 28 gm

Hence the mass of nitrogen which corresponds to 33.53 ml is $\frac{28g\times 33.53ml}{22400ml}=0.0419g$.

The percentage of nitrogen is $0.0419g\times \frac{100}{0.25g}=16.76$ %.

Hence, the correct option is $A$.