A) by adding\[Pb{{O}_{2}}\]into\[KMn{{O}_{4}}\]solution
B) by adding\[N{{a}_{2}}O\]to cold water
C) by adding\[Mn{{O}_{2}}\]to dil \[{{H}_{2}}S{{O}_{4}}\]
D) by passing\[C{{O}_{2}}\]into\[Ba{{O}_{2}}\]solution
Correct Answer: D
Solution :
In laboratory\[{{H}_{2}}{{O}_{2}}\]is prepared by passing\[C{{O}_{2}}\]into\[Ba{{O}_{2}}\]solution. \[Ba{{O}_{2}}+{{H}_{2}}O+C{{O}_{2}}\xrightarrow[{}]{{}}BaC{{O}_{3}}+{{H}_{2}}{{O}_{2}}\]
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