Question

The element silver (Ag) has two naturally occurring isotopes: 109Ag and 107Ag with a mass of 106.905 amu. Silver consists of 51.82 % Ag and has an average atomic mass of 107.868 amu. Calculate the mass of 109Ag

Answer 1

We know from average atomic mass formula,

Given the natural abundance % of Ag(109) = 48.16

                natural abundance % of Ag(107) = 51.82

Average atomic mass = 107.868 amu

Let mass of Ag(109) = $m_{1}$

     mass of Ag(107) = $m_{2}$ = 106.905 amu

On substituting all these values in given formula,

$107.868 = \frac{m_{1}*48.161+106.905*51.82  }{100}$

$48.161m_{1} = 10786.8- 5507.746 = 10236.[tex]</p><p>m_{1} =  \frac{10236.03}{48.161} =  212.538 amu$03[/tex]

Thus ,

The mass of Ag(109) =  212.538 amu

Answer 2

Answer:

Explanation:

The concept is to know about isotopes and average atomic mass