Question

Question #17fec

Answer 1

`3:1`

Explanation:

The trick here is the difference between the average relative atomic mass of chlorine, `A_r`, which you can take as `35.5`, and the relative isotopic masses of the two isotopes.

Since no information about the relative isotopic masses of the two isotopes was given, you can approximate them to be equal to the mass numbers.

Notice that the difference between the relative isotopic mass of `""^35"Cl"` and `""^37"Cl"` is

`37 - 35 = 2`

This means that you can split the difference between the two isotopes in `4` parts--`0.5` each--that correspond to a

`(100%)/4 = 25%`

difference in the percent abundance of the two isotopes.

You will have

`color(white)(a)`
`color(white)(aaaaaaaaaaaaaacolor(blue)("% abundance """^35"Cl"))`

`color(white)(aacolor(blue)(100%)aacolor(blue)(ul(75%))aaaacolor(blue)(50%)aaacolor(blue)(25%)aaaacolor(blue)(0%))`

`color(white)(aaacolor(black)(35)aaaaacolor(black)(ul(35.5))aaaaacolor(black)(36)aaaaacolor(black)(36.5)aaaaacolor(black)(37)) " "larr "A"_rcolor(white)(.)"of Cl"`

`color(white)(aaacolor(red)(0%)aaacolor(red)(ul(25%))aaaacolor(red)(50%)aaacolor(red)(75%)aaacolor(red)(100%))`
`color(white)(aaaaaaaaaaacolor(red)("% abundance """^37"Cl"))`

As you can see, a `50%-50%` split between the two isotopes would produce an average relative atomic mass of `36`.

Since the average relative atomic mass of chlorine is equal to `35.5`, you can say that the relative isotopic mass of the lighter isotope will contribute `color(blue)(75%)` and the relative isotopic mass of the heavier isotope will contribute `color(red)(25%)` to the average relative atomic mass of the element.

Therefore, the two isotopes exist in a ratio of

`(""^35"Cl")/(""^37"Cl") = (75color(red)(cancel(color(black)(%))))/(25color(red)(cancel(color(black)(%)))) = 3/1`