A) 0.1
B) 0.01
C) 0.001
D) 1
Correct Answer: A
Solution :
\[22\,4\,L\]of \[{{H}_{2}}=\frac{1}{22.4}\times 0.224\,mol\] \[=0.01\,mol\,of\,{{H}_{2}}\]
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