The element boron exists in nature as two isotopes: 10b has a mass of 10.0129 u, and 11b has a mass of 11.0093 u. the average atomic mass of boron is 10.81 u. calculate the relative abundance (as percentages) of the two boron isotopes.

The relative abundance of 10B = 0.20 (20%)

The relative abundance of 11B = 0.80 (80%)

Explanation:

Step 1: Data given

Boron has 2 natural isotopes

⇒ 10B has a mass of 10.0129 u

⇒ 11B has a mass of 11.0093 u

Average atomic mas of Boron = 10.81

Step 2:

10B has an abundance of X %

11B has an abundance of Y %

X+ Y = 1

X = 1 - Y

10.81 = 10.0129* (1 - Y) + 11.0093*Y

10.81 = 10.0129 - 10.0129Y + 11.0093Y

0.7971 = 0.9964Y

Y = 0.80

X = 1.0 - 0.80 = 0.20

10.0129*0.20 + 11.0093*0.80 = 2.00258 +8.8044 = 10.81

The relative abundance of 10B = 0.20 (20%)

The relative abundance of 11B = 0.80 (80%)

a

it is stable as it is in group 0 which have full outer shells of electrons