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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