A) 10%
B) 9%
C) 3%
D) 1%
Correct Answer: C
Solution :
: 1 litre of 10 volume\[{{H}_{2}}{{O}_{2}}\]gives 10 litre of oxygen at NTP 1 litre of 10 volume of\[{{H}_{2}}{{O}_{2}}\]gives\[\frac{10}{22.4}\]mol of oxygen From the equation, \[2{{H}_{2}}{{O}_{2}}\to 2{{H}_{2}}O+{{O}_{2}}\] mole of\[{{O}_{2}}=2\]mole of \[{{H}_{2}}{{O}_{2}}\] \[\frac{10}{22.4}\]mol of\[=\frac{2\times 10}{22.4}=0.893\]mole of \[{{H}_{2}}{{O}_{2}}\] \[\therefore \]Molarity of 10 vol sample of\[{{H}_{2}}{{O}_{2}}=0.893\] Strength of\[{{H}_{2}}{{O}_{2}}\]in 10 volume sample \[=0.893\times 34=30.4\text{ }g/L=3.04\text{ }%\].
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