Question

A sample of 0.125 g of an organic compound when analysed by Duma’s method yields 22.78 mL of nitrogen gas collected over KOH solution at 280 K and 759 mm Hg. The percentage of nitrogen in the given organic compound is____.(Nearest integer) 

Given : 1. The vapour pressure of water of 280 K is 14.2 mm Hg. 

            2. R = 0.082 L atm K^–1 mol^–1

Answer 1

\(\begin{array}{l}P_{original} = 759 – 14.2 = 744.8 ~\text{mmHg}\end{array}\)

\(\begin{array}{l}n_{N_2}=\frac{744.8\times22.78}{760\times0.0821\times280\times1000}\\ = 0.000971 ~\text{mol}\end{array}\)

\(Mass \;of N_2 = 0.02719 \;gm\)

\(\begin{array}{l}\text{Percentage of nitrogen} =\frac{0.0271}{0.125}\times100=21.75\simeq 22 \end{array}\)