Question

(b) An element X is available in the form of two isotopes with same atomic no 15. The isotope
with mass no 31 and other with mass no 33, present in the sample is (75.8%) and(24.2%)
respectively. Calculate the average atomic mass of element X. [S1 and 13X
Write the three equation of motion. Obtain a relation for the
15.​

Answer 1

Explanation:

Given Natural abundance of lighter element =20%

⇒

56

X=20%

∴

58

X=80%

Average atomic mass =(56×0.2)+(58×0.8)=57.6