Question
In Duma's method $$0.52\,g$$ of an organic compound on combustion gave $$68.6\,mL\,{N_2}$$ at $${27^ \circ }C$$ and $$756\,mm$$ pressure. What is the percentage of nitrogen in the compound?
A.
$$12.22\% $$
B.
$$14.93\% $$
C.
$$15.84\% $$
D.
$$16.23\% $$
Answer :
$$14.93\% $$
Solution :
$$\eqalign{
& {V_1} = 68.6\,mL,{P_1} = 756\,mm,{T_1} = 300\,K \cr
& {V_2} = ?,{P_2} = 760\,mm,{T_2} = 273\,K \cr
& \frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}} \cr
& {\text{At}}\,NTP,\,vol.\,{\text{of}}\,{N_2},{V_2} = \frac{{{P_1}{V_1}}}{{{T_1}}} \cdot \frac{{{T_2}}}{{{P_2}}} \cr
& = \frac{{756 \times 68.6}}{{300}} \times \frac{{273}}{{760}} \cr
& = 62.09\,mL \cr} $$
Percentage of nitrogen in organic compound
$$\eqalign{
& = \frac{{28}}{{22400}} \times \frac{{{V_2}}}{w} \times 100 \cr
& = \frac{{28}}{{22400}} \times \frac{{62.09}}{{0.52}} \times 100 \cr
& = 14.93\% \cr} $$