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