The correct answer for the question that is being presented above is this one:
Boron consists of two isotopes, boron-10 and boron-11.
Given that its relative atomic mass is 10.2
Find the abundance of each isotope.
Let y/100 = the abundance of copper-10
and (100 - y)/100 = the abundance of copper-11
10.2 = (y/100 x 10) + [(100 - y)/100 x 11]
10.2 = 10y/100 + 1100/100 - 11y/100
1020 = 10y + 1100 - 11y
-80 = -y
y = 80
Abundance of boron-10 = 10/100 = 10%
Abundance of boron-11 = 100 - 10 = 90%
