A) 10
B) 3
C) 24
D) 5
Correct Answer: A
Solution :
[a] Milli equivalents of \[{{H}_{2}}S{{O}_{4}}\] \[=60\times \frac{M\times 2}{10}=12\] Milli equivalents of \[NaOH=20\times \frac{M}{10}=2\] Milli equivalent of \[NH=12-2=10\] % of nitrogen \[=\frac{1.4\times \left( N\times V \right)N{{H}_{3}}}{\left( Wt.\text{ }of\text{ }organic\text{ }compound \right)}\] \[\frac{1.4\times 10}{1.4}=10\]You need to login to perform this action.
You will be redirected in
3 sec