Let x = fractional abundance Co60
1-x = fractional abundance Co59
(Note that as percentages that will be x% and 100%-x%). Then
x(atomic mass Co60) + (1-x)(atomic mass Co59) = 58.8901
Solve for x and 1-x will give you the fractinal abundances of Co59 and Co60. Then fraction * 1.8155g = grams Co60 or Co59 depending upon which fraction you use. Post your work if you get stuck.
Naturally occurring cobalt consists of only one isotope, cobalt59, whose relative atomic mass is 58.9332. A synthetic radioactive isotope of cobalt60, relative atomic mass 59.9338, is used in radiation therapy for cancer. A 1.8155 sample of cobalt has an apparent "atomic mass" of 58.9901.
FIND THE MASS OF COBALT60 IN THIS SAMPLE.
I DON'T KNOW WHERE TO START, I TRIED DOING CONVERSION FACTORS BUT I CANT GET A REASONABLE MASS.
HELP?
5 answers
I get....
(x)(59.9338) +(1-x)(58.9332) =(58.9901)
59.9338x+58.9332-58.9332x=58.9901
1.0006x=0.0560
x=0.0569
cobalt60 abundance=0.0569%
cobalt59 abundance=0.9431%
0.0569/0.9431 X 1.8155g = .10953grams
Is that right?? It looks like i did it wrong.
(x)(59.9338) +(1-x)(58.9332) =(58.9901)
59.9338x+58.9332-58.9332x=58.9901
1.0006x=0.0560
x=0.0569
cobalt60 abundance=0.0569%
cobalt59 abundance=0.9431%
0.0569/0.9431 X 1.8155g = .10953grams
Is that right?? It looks like i did it wrong.
We get...
OH wait, do we do
0.0569 X 1.8155g = 0.103g ?
so cobalt60=0.103g
OH wait, do we do
0.0569 X 1.8155g = 0.103g ?
so cobalt60=0.103g
Thanks!
Everything looks ok except where you converted to percent (which the problem didn't ask for). x = 0.0560 = Co60 abundance which is 5.60% if yu want to convert to percent.
1 answer