Given:
m1 = 10.013 amu
m2 = 11.009 amu
m = 10.81 amu
use:
atomic mass = sum of (mass of isotope * abundance) / 100
m = (m1*A1 + m2*A2)/100
10.81 = ( 10.013*A1 + 11.009*(100-A1))/100
10.81 = ( 10.013*A1 + 1100.9- 11.009*A1))/100
A1 = 20.0 %
So,
A2 = 100 - A1
= 100 - 20
= 80 %
Clearly 2nd isotope is more abundant
Answer: boron-11
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