A) \[{{H}_{2}}C=CH-C\equiv CH\]
B) \[HC\equiv C-C{{H}_{2}}-C\equiv CH\]
C) \[NaOH\]
D) \[{{104}^{o}}40\]
Correct Answer: A
Solution :
Perhydrol means 30% solution of \[{{H}_{2}}{{O}_{2}}\]. \[{{H}_{2}}{{O}_{2}}\] decomposes as \[2{{H}_{2}}{{O}_{2}}\xrightarrow{{}}2{{H}_{2}}O+{{O}_{2}}\] Volume strength of 30% \[{{H}_{2}}{{O}_{2}}\] solution is 100 that means \[1mL\] of this solution on decomposition gives \[100mL\] oxygen. \[S{{O}_{2}}+\frac{1}{2}{{O}_{2}}\xrightarrow{{}}S{{O}_{3}}\] \[\begin{matrix} 1L \\ 2L \\ \end{matrix}\,\,\,\,\,\,\,\,\,\,\,\,\,\begin{matrix} \frac{1}{2}L \\ 1L \\ \end{matrix}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\begin{matrix} 1L \\ 2L \\ \end{matrix}\] Since, \[100mL\] of oxygen is obtained by \[=1mL\]of \[{{H}_{2}}{{O}_{2}}\] \[\therefore \] \[1000mL\] of oxygen will be obtained by \[=\frac{1}{100}\times 1000mL\] of \[{{H}_{2}}{{O}_{2}}\] \[=10mL\] of \[{{H}_{2}}{{O}_{2}}\]
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