Respuesta :
Answer:
There are [tex]1.0161\times 10^{21} [/tex] atoms of 13-C present in a 2.05 grams sample of carbon dioxide.
Explanation:
Percentage of 12-C isotope = 98.93%
Percentage of 13-C isotope = 1.07%
Mass of the sample of carbon dioxide = 2.05 g
Mass of 13-C in the sample = 1.07% of 2.05 g
[tex]=\frac{1.07}{100}\times 2.05 g=0.021935 g[/tex]
Molar mass of 13-C isotope = 13 g/mol
Moles of 13-C = [tex]\frac{0.021935 g}{13 g/mol}=0.001687 mol[/tex]
[tex]1 mol = 6.022\times 10^{23}[/tex] atoms/ molecules
Atoms of 13-C in 0.001687 moles:
[tex]0.001687 \times 6.022\times 10^{23}=1.0161\times 10^{21} [/tex] atoms
There are [tex]1.0161\times 10^{21} [/tex] atoms of 13-C present in a 2.05 grams sample of carbon dioxide.