A) \[0.085g\]
B) \[0.850g\]
C) \[8.500g\]
D) \[80.500g\]
Correct Answer: A
Solution :
: Mass of \[22400c{{m}^{3}}\]of \[N{{H}_{3}}=17\] Mass of \[112c{{m}^{3}}\]of \[N{{H}_{3}}=\frac{17\times 112}{22400}=0.085g\]
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