A) \[112\,ml\]
B) \[224\,ml\]
C) \[56\,ml\]
D) \[336\,ml\]
Correct Answer: B
Solution :
We know that \[2{{H}_{2}}{{O}_{2}}\xrightarrow{{}}\,2{{H}_{2}}O+{{O}_{2}}\] \[2\times 34g\] \[22400\,ml\] \[\because \] \[2\times 34\,gm=68\,gm\] of \[{{H}_{2}}{{O}_{2}}\] liberates \[22400\,ml\,\,{{O}_{2}}\] at STP \[\therefore \] \[.68\,gm\] of\[{{H}_{2}}{{O}_{2}}\] liberates \[=\frac{.68\times 22400}{68}=224\,ml\]
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