A) 24
B) 5
C) 10
D) 3
Correct Answer: C
Solution :
Mili equivalents of \[{{H}_{2}}S{{O}_{4}}=60\times \frac{M\times 2}{10}=12\] Mili equivalents of \[NaOH=20\times \frac{M}{10}=2\] Mili equivalents of \[N{{H}_{3}}=12-2=10\] % of Nitrogen \[=\frac{1.4\times \left( N\times V \right)N{{H}_{3}}}{W}\] \[=\frac{1.4\times 10}{1.4}=10\]You need to login to perform this action.
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