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QUESTION

# The element rubidium (atomic mass 85.5) consists of two isotopes, one mass number 85 (Rb-85) and the other mass number 87 (Rb-87). Which one of the following statements about the isotopes of rubidium is correct?

Alternative approach.

Here's another approach to use in order to find the abundances of the two .

As you know, each isotope will contribute to the average of rubidium in proportion to their abundance.

color(blue)(|bar(ul(color(white)(a/a)"avg. atomic mass" = sum_i i xx "abundance"_icolor(white)(a/a)|)))

Here i represents the atomic mass of an isotope i. This equation uses decimal abundance, which is simply percent abundance divided by 100

color(blue)(|bar(ul(color(white)(a/a)"decimal abundance" = "percent abundance"/100color(white)(a/a)|)))

For example, if an isotopes has a 13% percent abundance, it will have a

"decimal abundance" = 13/100 = 0.13

So, you know that your element has two stable isotopes, ""^85"Rb" and ""^87"Rb". If you take x to be the decimal abundance of ""^85"Rb", you can say that the decimal abundance of ""^87"Rb" will be 1-x.

This is the case because the abundances of the two must add up to give 100%, or 1 as a decimal abundance.

You know that the average atomic mass of rubidium is "85.5 u", and that the two isotopes have atomic masses equal to "85 u" and "87 u", respectively.

The equation will thus take the form

85.5 color(red)(cancel(color(black)("u"))) = 85color(red)(cancel(color(black)("u"))) xx x + 87color(red)(cancel(color(black)("u"))) xx (1-x)

85.5 = 85x + 87 - 87x

Rearrange to solve for x

87x - 85x = 87 - 85.5

2x = 1.5 implies x = 0.75

So, the abundances of the two isotopes will be

"For """^85"Rb: " overbrace(0.75)^(color(purple)("decimal abundance")) = overbrace(75%)^(color(green)("percent abundance"))

"For """^87"Rb: " 1 - 0.75 = overbrace(0.25)^(color(purple)("decimal abundance")) = overbrace(25%)^(color(green)("percent abundance"))

Once again, this shows that ""^85"Rb" is three times as abundant as ""^87"Rb".

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