• # question_answer The volume of oxygen liberated from $0.68gm$ of ${{H}_{2}}{{O}_{2}}$ is [Pb. PMT 2004] A) $112\,ml$ B) $224\,ml$ C) $56\,ml$ D) $336\,ml$

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 $=\frac{.68\times 22400}{68}=224\,ml$