Answer:
Option A is correct = 157.2 amu
Explanation:
Given data:
Percentage abundance of Zr-156 = 84%
Atomic mass of Zr-156 = 156.8 amu
Percentage abundance of Zr-160 = 16%
Atomic mass of Zr-160 = 159.4 amu
Average atomic of Zr on Mars = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (84×156.8)+(16×159.4) /100
Average atomic mass = 13171.2 + 2550.4 / 100
Average atomic mass = 15721.6 / 100
Average atomic mass = 157.2 amu.