Answer:
Mass of the organic compound =
0.50 g
Unused acid required
= 60 mL of 0.5 M \[NaOH\]
=30 ML of 1 M\[NaOH\]
Volume of
acid taken = 50 mL of 0.5 M \[{{H}_{2}}S{{O}_{4}}\]
= 25 mL of 1 M \[{{H}_{2}}S{{O}_{4}}\]
Now 1 mole
of \[{{H}_{2}}S{{O}_{4}}\] neutralizes 2 moles of \[NaOH\]
\[({{H}_{2}}S{{O}_{4}}+2NaOH\to
N{{a}_{2}}S{{O}_{4}}+2{{H}_{2}}O)\]
So, 30 mL of
1M \[NaOH\] = 15 mL of 1 M \[{{H}_{2}}S{{O}_{4}}\]
So, volume
of acid used by ammonia
= 25 - 15 = 10 mL
% of
nitrogen = \[\frac{1.4\times {{N}_{1}}\times V}{W}=\frac{1.4\times 1\times 2\times
10}{0.5}\]
= 56%
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