A) 8.4 L
B) 4.2 L
C) 16.8 L
D) 5.2 L
Correct Answer: A
Solution :
\[Strength=Normality\times equivalent\text{ }weight\] \[=1.5\times 17\] \[=25.5\] Mol. wt. of \[{{H}_{2}}{{O}_{2}}=1\times 2+16\times 2=2+32=34\] \[\underset{2(34)=68g}{\mathop{2{{H}_{2}}{{O}_{2}}}}\,\xrightarrow[{}]{{}}2{{H}_{2}}O+{{O}_{2}}\] \[\because \]68 g of\[{{H}_{2}}{{O}_{2}}\]liberates = 22.4 L of\[{{O}_{2}}\]at NTP \[\therefore \]25.5 g of\[{{H}_{2}}{{O}_{2}}\]liberates\[=\frac{22.4}{68}\times 25.5\] = 8.4 L of\[{{O}_{2}}\]at NTP
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