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