A) 30%
B) 10%
C) 6%
D) 3%
Correct Answer: C
Solution :
20% volume\[{{H}_{2}}{{O}_{2}}\]means, 20 ml \[{{O}_{2}}\]is obtained from the decomposition of 1 ml\[{{H}_{2}}{{O}_{2}}\] at S.T.P. \[\underset{2\,\times \,34\,=\,68}{\mathop{2{{H}_{2}}{{O}_{2}}}}\,\xrightarrow{{}}2{{H}_{2}}O+\underset{\begin{smallmatrix} 1\,\,\text{mole}\,\text{=}\,\text{22}\text{.4}\,\,\text{litre}\,\,\text{at}\,\text{S}\text{.T}\text{.P}\text{.} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{=}\,\text{22400}\,\,\text{ml} \end{smallmatrix}}{\mathop{{{O}_{2}}}}\,\] \[\because \] 22400 ml \[{{O}_{2}}\]is obtained by the decomposition of \[=68g\,{{H}_{2}}{{O}_{2}}\] \[\therefore \] 20 ml \[{{O}_{2}}\] will be obtained by the decomposition of \[=\frac{68\times 20}{22400}g\,{{H}_{2}}{{O}_{2}}\] Hence, the strength of \[{{H}_{2}}{{O}_{2}}\] solution \[=\frac{68\times 20}{22400}g\text{/}ml\] \[\therefore \] \[%\] strength\[=\frac{68\times 20\times 100}{22400}=6.07%\]
You need to login to perform this action.
You will be redirected in
3 sec
