A) \[112\,\,ml\]
B) \[224\,\,ml\]
C) \[56\,\,ml\]
D) \[336\,\,ml\]
Correct Answer: B
Solution :
We know that\[\underset{2\times 34\,\,g}{\mathop{2{{H}_{2}}{{O}_{2}}}}\,\xrightarrow{{}}2{{H}_{2}}O\underset{22400\,\,ml}{\mathop{+\,{{O}_{2}}}}\,\] \[\because \]\[2\times 34\,\,gm=68\,\,gm\]of\[{{H}_{2}}{{O}_{2}}\]liberates\[22400\,\,ml\,\,{{O}_{2}}\]at STP \[\therefore \]\[0.68\,\,gm\]of\[{{H}_{2}}{{O}_{2}}\]liberates \[=\frac{0.68\times 22400}{68}=224\,\,ml\]
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