Relative abundance — task. Science TNSB Mentoring, Class 10 Crash course.

Write the correct answer:

Identify the % relative abundance of \(C-12\) and \(C-13\), if its average atomic mass is \(12.01\) \(amu\).

Let the % of the isotope \(C-12=x\)

Then, the % of the isotope \(C-13=100-x\)

\begin{array}{l} Average atomic mass = \frac{12 \times x + 13 (100 - x)}{100} \\ 12.01 = \frac{A}{100} \\ 12.01 = \frac{B}{100} \\ x = ? \end{array}

Where,

A \(=\)

B \(=\)

x \(=\)

% abundance of \(C-12=\)

% abundance of \(C-13=\)

Did you find an error?

7. Relative abundance