A) 1.0
B) 0.5
C) 11.2
D) 1.12
E) 0.75
Correct Answer: A
Solution :
\[1\text{ }mL\text{ }{{H}_{2}}{{O}_{2}}\]solution gives\[11.2\text{ }mL\text{ }{{\text{O}}_{2}}\]at NTP \[\therefore \]\[100\text{ }mL\text{ }{{H}_{2}}{{O}_{2}}\]solution gives \[{{O}_{2}}=100\times 11.2\] \[=1120.0\text{ }mL\,{{O}_{2}}\text{ }at\text{ }NTP\] \[{{H}_{2}}{{O}_{2}}\]decomposes as \[2{{H}_{2}}{{O}_{2}}(l)\xrightarrow{{}}2{{H}_{2}}{{O}_{2}}(l)+{{O}_{2}}(g)\] \[\because \]\[22400\text{ }mL\text{ }{{\text{O}}_{2}}\]at NTP is obtained from 68g \[{{H}_{2}}{{O}_{2}}\] \[\therefore \]\[1\text{ }mL\text{ }{{\text{O}}_{2}}\]at NTP is obtained from \[=\frac{68}{22400}g\,{{H}_{2}}{{O}_{2}}\] \[\therefore \]\[1120\text{ }mL\,{{O}_{2}}\]at NTP is obtained from \[=\frac{68}{22400}\times 1120\] \[=34\,g\] \[w=\frac{M\times m\times v}{1000}\] \[M=1.0\]
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