Boron has two naturally occurring isotopes, 10b and 11b, which have masses 10.0129 and 11.0093 amu, respectively. given the average atomic mass of boron (10.81 amu), determine the percent abundance of each isotope. (a) 50%10b, 50%11b (b) 20%10b, 80%11b (c) 98%10b, 2% 11b (d) 93%10b, 7%11b (e) 22%10b, 78%11b

The correct answer is option b.

Explanation:

Let the fractional abundance of B-10 be x

And fractional abundance of B-11 be (1- x)

Average atomic mass of boron = 10.81 amu

Atomic mas of B-10 = 10.0129 amu

Atomic mas of B-11 = 11.0093 amu

To determine an average atomic mass of an element we use::

x = 0.200 = 0.200 × 100= 20%

(1-x) = 0.800 =0.800 × 100= 80%

The percent abundance of isotopes B-10 and B-11 is 20% and 80% respectively.