Question

Naturally occurring boron consists of two isotopes whose Z are 10 and 11 the atomic mass of natural boron is 10.80 calculate percentage of each isotope in natural boron

Answer 1

Answer:

percentage are 20 and 80

Explanation:

consider percentage of each isotope be x and 100‐x

so,

10.80 = 10x + 11(100‐x)/x+100‐x

10.80 = 10x + 1100 ‐11x /100

10.80×100= 1100 ‐ x

1080 ‐ 1100 = ‐x

‐20 = ‐x

x = 20

100 - x = 80

therefore percentage are 20% and 80%