A) 35
B) 47
C) 10
D) 25
Correct Answer: C
Solution :
% of \[C=\frac{12}{44}\times \frac{Weight\,\,of\,\,C{{O}_{2}}}{Weight\,\,of\,\,compound}\times 100\] \[=\frac{12}{44}\times \frac{0.66}{0.20}\times 100=90%\] Thus, % of \[H=100-%\,\,of\,\,C\] \[=100-90=10%\]
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