A) 3.00%
B) 4.045%
C) 2.509%
D) 3.035%
Correct Answer: D
Solution :
\[[{{H}_{2}}{{O}_{2}}\to {{H}_{2}}O+\frac{1}{2}{{O}_{2}}]\times 2\] \[\underset{68\,g}{\mathop{2{{H}_{2}}{{O}_{2}}}}\,\to 2{{H}_{2}}O+{{O}_{2}}\] 22.4 litre at N.T.P. \[\because \] 22.4 litre \[{{O}_{2}}\] at N.T.P. obtained by 68 gm of \[{{H}_{2}}{{O}_{2}}\] \ 10 litre \[{{O}_{2}}\] at N.T.P. obtained by \[\frac{68}{22.4}\times 10=30.35\,gm/litre\] \ 1000 ml \[{{O}_{2}}\] at N.T.P. obtained by = 30.35 gm \ 100 ml \[{{O}_{2}}\] at N.T.P. obtained by \[=\frac{30.35}{1000}\times 100=3.035%\]
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