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 300K is 25mm, the percentage of nitrogen in the compound is :

• A. 15.76
• B. 17.36
• C. 18.20
• D. 16.76

Answer 1

The correct option is D

16.76

0.25 g 40 mL N2 at 300K, 725 mm pressure

Aq. tension at 300 K is 25 mm

725 - 25 = 700 mm

Temp. 300 K , Mass of the sub 0.25 g , Vol. of moist nitrogen = 40 mL

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

$\therefore$ $V_2$ = $\frac{P_1{V}_1}{T_1}$ $\times$ $\frac{T_2}{P_2}$ = $\frac{700\times 40\times 273}{300\times 760}$ = $\frac{7644000}{228000}$ = 33.52mL

% of nitrogen

22400 mL of nitrogen at S.T.P weighs = 28 g

33.52 mL of nitrogen at S.T.P. weighs

= $\frac{28\times 33.52}{22400}$ = $\frac{938.56}{22400}$ = 0.0419g

Percentage of nitrogen in org. compound = $\frac{0.0419}{0.25}$ $\times$ 100 = 16.76%