The mass $\%$ of carbon in $\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_6$ is
The mass $\%$ of carbon in $\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_6$ is
- 57.95
- 62.35
- $73.45^{\circ}$
- 76.85
Solution
Molecular mass of
$
\begin{aligned}
\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_6= & (12 \times 57)+(110 \times 1)+(6 \times 16) \\
=684+110+ & 96=890 \\
\text { Mass } \% \text { of } \mathrm{C} & =\frac{\text { Mass of } \mathrm{C}}{\text { Molecular mass of compound }} \times 100 \\
& =\frac{684}{890} \times 100=76.85 \%
\end{aligned}
$
Asked in: AP EAMCET 2022 (06 Jul Shift 2)
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