Phosphorus

Let the percentage abundance of the lighter isotope (

%5Cmathrm%7B%5E%7B10%7DB%7D

) be

x%20%5C%25

. Then the percentage abundance of the heavier isotope (

%5Cmathrm%7B%5E%7B11%7DB%7D

) will be

(100%20-%20x)%5C%25

. The average atomic mass of boron is given by:

%5Ctext%7BAverage%20atomic%20mass%7D%20%3D%20%5Cfrac%7B(x%20%5Ccdot%2010)%20%2B%20((100%20-%20x)%20%5Ccdot%2011)%7D%7B100%7D.

Substitute the given average atomic mass

10.81

10.81%20%3D%20%5Cfrac%7B(x%20%5Ccdot%2010)%20%2B%20((100%20-%20x)%20%5Ccdot%2011)%7D%7B100%7D.

Simplify:

10.81%20%3D%20%5Cfrac%7B10x%20%2B%201100%20-%2011x%7D%7B100%7D.

Combine like terms:

10.81%20%3D%20%5Cfrac%7B1100%20-%20x%7D%7B100%7D.

Multiply through by 100:

1081%20%3D%201100%20-%20x.

Solve for

x

x%20%3D%201100%20-%201081%20%3D%2019.

Thus, the abundance of the lighter isotope is

%5Cmathbf%7B19%5C%25%7D

Chapter Questions
1 MCQs