A) 30%
B) 3%
C) 1%
D) 10%
Correct Answer: B
Solution :
10 volume \[{{\text{H}}_{\,\text{2}}}{{\text{O}}_{\text{2}}}\text{''}\]means 1 mL of its solution on decomposition at NTP, give 10 mL oxygen gas. Volume of oxygen formed from 100 mL of solution at\[\text{ }\!\!~\!\!\text{ NTP}\,\text{=}\,\text{1000 mL}\] \[\underset{\text{2}\,\text{ }\!\!\times\!\!\text{ }\,\text{34g}}{\mathop{\underset{\text{2}\,\text{mol}}{\mathop{\text{2}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}}}\,}}\,\xrightarrow{{}}\text{2}{{\text{H}}_{\text{2}}}\text{O}\,\text{+}\underset{\text{22400}\,\text{mL}}{\mathop{\underset{\text{1}\,\text{mol}}{\mathop{{{\text{O}}_{\text{2}}}}}\,}}\,\] \[\because \] \[\text{22400}\,\text{mL}\,{{\text{O}}_{\text{2}}}\]formed at NTP by decomposition of \[\text{68}\,\text{g}\,{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\text{.}\] \[\ \therefore \text{ }\] \[\text{1}\,\text{mL}\,{{\text{O}}_{\text{2}}}\]formed at NTP from\[\frac{68}{22400}g\]of \[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\] \[\ \therefore \text{ }\]\[\text{1000 mL }{{\text{O}}_{\text{2}}}\] formed at NTP from \[\frac{68\times 1000}{22400}g\,{{H}_{2}}{{O}_{2}}=3.035\,g\,{{H}_{2}}{{O}_{2}}\] So, concentration of 10 volume \[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\text{''}\] \[=3.0%\]approximately
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