Silver has two naturally occurring isotopes with the following isotopic masses: 47ag 107 – 106.90509 47ag 109 – 108.9047 the average atomic mass of silver is 107.8682 amu. the abundance of the lighter of the two isotopes is 3/1

a) 0.2422b) 0.4816c) 0.5184d) 0.7578e) 0.9047

c) .51835

Explanation:

Let the relative abundance of the lighter of the two isotopes be X we have

Then the relative abundance of the heavier isotope is then (1-X)

Whereby we have that in nature the amount of the lighter silver found in proportion is X and the heavier isotope of silver is present as (1-X) proportion in nature.

To calculate the relative atomic mass of silver, we have

(Mass of light weight silver)×X + (mass of heavier isotope of silver×(1-X) = relative atomic mass of silver

106.90509(X) + 108.9047(1-X)

108.9-108.9(x)+106.9(x) = 107.87

-2x-1.03 = 0.517450902926

Closest answer is c

c) .5184

The relative atomic mass of isotopes is the weighted average by the mole-fraction of abundance of these isotopes which gives the atomic weight that is listed for that element on the periodic table.

i'm pretty sure its b i might be wrong though!