A) 30%
B) 6%
C) 3%
D) 10%
Correct Answer: B
Solution :
20 volumes of \[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\]means 20 L of oxygen is obtained at STP by the decomposition of 1 L \[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\]solution. \[\underset{\begin{smallmatrix} \text{2}\,\text{mol} \\ \text{68}\,\text{g} \end{smallmatrix}}{\mathop{2{{H}_{2}}{{O}_{2}}}}\,\xrightarrow{{}}2{{H}_{2}}O+\underset{\begin{smallmatrix} \,\,\,\,\,\,\,\,\,\,\text{1}\,\text{mol} \\ \text{22}\text{.4}\,\text{L}\,\text{at}\,\text{STP} \end{smallmatrix}}{\mathop{{{O}_{2}}}}\,\] 22.4 L oxygen is obtained from \[\text{68}\,\text{g}\,{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\text{.}\] \[\therefore \] 20 L oxygen is obtained from \[=\frac{68}{22.4}\times 20=60.7g\,{{H}_{2}}{{O}_{2}}\] 1000 mL of\[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\] solution contains 60.7 g \[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\text{.}\] \[\therefore \] 100 mL \[\text{(}\approx \text{100}\,\text{g)}\] solution will contain \[=\frac{60.7}{1000}\times 100g\,{{H}_{2}}{{O}_{2}}\] \[=6.07g\,{{H}_{2}}{{O}_{2}}\] \[\therefore \]strength of \[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\]solution \[=6.07%\]
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