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The mass $\%$ of carbon in $\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_6$ is
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Verified Answer
The correct answer is:
76.85
Molecular mass of
$$
\begin{aligned}
\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_6= & (12 \times 57)+(110 \times 1)+(6 \times 16) \\
=684+110+ & 96=890 \\
\text { Mass } \% \text { of } \mathrm{C} & =\frac{\text { Mass of } \mathrm{C}}{\text { Molecular mass of compound }} \times 100 \\
& =\frac{684}{890} \times 100=76.85 \%
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_6= & (12 \times 57)+(110 \times 1)+(6 \times 16) \\
=684+110+ & 96=890 \\
\text { Mass } \% \text { of } \mathrm{C} & =\frac{\text { Mass of } \mathrm{C}}{\text { Molecular mass of compound }} \times 100 \\
& =\frac{684}{890} \times 100=76.85 \%
\end{aligned}
$$
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