A) 3
B) 6
C) 9
D) 12
Correct Answer: A
Solution :
\[\text{ }\!\!\!\!\text{ 10}\,\text{V}\,{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\,\text{ }\!\!\!\!\text{ }\]means 1 L of this solution will produce \[\text{10}\,\text{L}\,{{\text{O}}_{2}}\] at STP. \[\underset{68\,g}{\mathop{2{{H}_{2}}{{O}_{2}}\,}}\,\xrightarrow{{}}2{{H}_{2}}O+\underset{22.4\,L\,\text{at}\,\text{STP}}{\mathop{{{O}_{2}}}}\,\] \[\because \]\[\text{22}\text{.4}\,\text{L}\]of \[{{\text{O}}_{2}}\]obtained from \[{{\text{H}}_{2}}{{O}_{2}}=68\,g\] \[\therefore \]\[10\,L\]of \[{{\text{O}}_{\text{2}}}\]will be obtained from \[{{H}_{2}}{{O}_{2}}=\frac{68}{22.4}\times 10=30.36\,g\] \[\therefore \]1000 mL of the given solution contains 30.36 g \[{{H}_{2}}{{O}_{2}}\]and 100 mL of the given solution contains \[\frac{30.36\times 100}{1000}=3.03\,g\,{{H}_{2}}{{O}_{2}}\] Thus, % strength of \[{{H}_{2}}{{O}_{2}}\]is \[3.03\,(\approx 3).\]
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