$0.765 \mathrm{~g}$ of an acid gives $0.535 \mathrm{~g}$ of $\mathrm{CO}_2$ and $0.13 \mathrm{~g}…
$0.765 \mathrm{~g}$ of an acid gives $0.535 \mathrm{~g}$ of $\mathrm{CO}_2$ and $0.13 \mathrm{~g} \mathrm{of}_2 \mathrm{O}$, then ratio of percentage of $\mathrm{C}$ and $\mathrm{H}$ is
19 : 2
18 : 11
20 : 17
1 : 7
Solution
Given, mass of $\mathrm{CO}_2=0.535$
Mass of compound $=0.765 \mathrm{~g}$
Now, percentage of carbon
$
\begin{aligned}
& =\frac{\text { weight }}{\text { molecular wt. }} \times \frac{\text { mass of } \mathrm{CO}_2}{\text { mass of compound }} \times 100 \\
& =\frac{12}{44} \times \frac{0.535}{0.765} \times 100 \\
\mathrm{C} & =19.07 \% \\
& =19 \%
\end{aligned}
$
Similarly, mass of $\mathrm{H}_2 \mathrm{O}=0.13 \mathrm{~g}$
mass of compound $=0.765 \mathrm{~g}$
Now, percentage of hydrogen $=\frac{2}{18} \times \frac{0.138}{0.765} \times 100$
$\% \mathrm{H}=2.00 \%$
$\therefore$ Ratio of $\mathrm{C}$ and $\mathrm{H}$ is $19: 2$
Hence, the ratio of percentage of $\mathrm{C}$ and $\mathrm{H}$ is $19: 2$