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Natural abundances of ${ }^{12} \mathrm{C}$ and ${ }^{13} \mathrm{C}$ isotopes of carbon are $99 \%$ and $1 \%$, respectively. Assuming they only contributes to the mol. wt. of $\mathrm{C}_{2} \mathrm{F}_{4}$, the percentage of $\mathrm{C}_{2} \mathrm{F}_{4}$ having a molecular mass of $101$ is
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$1.98$
Percentage of $\mathrm{C}_{2} \mathrm{F}_{4}$ of Molar mass $100=\frac{1}{100} \times \frac{1}{100} \times 100=0.01 \%$
Percentage of $\mathrm{C}_{2} \mathrm{F}_{4}$ of Molar mass $102=\frac{99}{100} \times \frac{99}{100} \times 100=98.01 \%$
Percentage of $\mathrm{C}_{2} \mathrm{F}_{4}$ of Molar mass $101=100-(0.01+98.01)=1.98 \%$
Percentage of $\mathrm{C}_{2} \mathrm{F}_{4}$ of Molar mass $102=\frac{99}{100} \times \frac{99}{100} \times 100=98.01 \%$
Percentage of $\mathrm{C}_{2} \mathrm{F}_{4}$ of Molar mass $101=100-(0.01+98.01)=1.98 \%$
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