Ordinary chlorine gas has 2 isotopes 35-Cl-17 and 37-Cl-17 in the Ratio 3:1 Calculate the relative atomic mass [atomic weight] of chlorine. (With Working)
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Let us consider 4 atoms of a random Cl sample.
By the given ratio, we have 3 35-CL-17 atoms and 1 37-Cl-17 atom, among these 4 atoms.
So, total atomic mass of 4 atoms= 35X3+37=142
So, relative atomic mass for 1 atom= (142/4)=35.5
By the given ratio, we have 3 35-CL-17 atoms and 1 37-Cl-17 atom, among these 4 atoms.
So, total atomic mass of 4 atoms= 35X3+37=142
So, relative atomic mass for 1 atom= (142/4)=35.5
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Ar(Cl) = (abundance of 35-Cl)*(mass of 35-Cl) + (abundance of 37-Cl)*(mass of 37-Cl)
Ar(Cl) = 0.75*35 + 0.25*37 = 26.25 + 9.25 = 35.5
(Note: The actual Ar(Cl) is 35.453, which is close enough to the result we obtained.)
Ar(Cl) = 0.75*35 + 0.25*37 = 26.25 + 9.25 = 35.5
(Note: The actual Ar(Cl) is 35.453, which is close enough to the result we obtained.)