A) 30%
B) 6%
C) 3%
D) 10%
Correct Answer: B
Solution :
[b] \[\therefore 22.4\] litre \[{{O}_{2}}\] at N.T.P. obtained by 68 gm of \[{{H}_{2}}{{O}_{2}}\] \[\therefore \] 1 litre \[{{O}_{2}}\] at N.T.P. obtained by \[\frac{68}{22.4}gm\] of \[{{H}_{2}}{{O}_{2}}\] \[\therefore \text{ }20\] litre \[{{O}_{2}}\] at N.T.P. obtained by \[\frac{68}{22.4}\times 20gm\,of\,{{H}_{2}}{{O}_{2}}=60.71gm\,of\,{{H}_{2}}{{\text{O}}_{2}}\] \[\therefore 1000\,ml\,{{O}_{2}}\] at N.T.P. obtained by \[=60.71\,gm\,of\,{{H}_{2}}{{\text{O}}_{2}}\] \[\therefore 100\,ml\,{{O}_{2}}\] at N.T.P. obtained by \[=\frac{60.71}{1000}\times 100=6.071%\]%
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