Explain Isobar ? An element occurrence in nature as X & X is 3:1 ratio, find its average atomic mass & confirm the name of element ?[1.5 + 1 + 0.5]
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Average atomic mass = f1M1 + f2M2 +… + fnMn where f is the fraction representing the natural abundance of the isotope and M is the mass number (weight) of the isotope.
In the above question the ratio is M:N.
Mass the with natural abundance M=10 u
Mass the with natural abundance N=11 u
Now just put the values in the formula and you get the answer.
[(10M)/(M+N)]+[(11N)/(M+N)] = 10.8 [STEP-1]
[(10M+11N)/(M+N)] = 10.8 [STEP-2]
10M+11N = 10.8M +10.8N [STEP-3] [Transposing (M+N) to R.H.S.]
Therefore
0.2N = 0.8M [STEP-4]
Hence M/N = 0.2/0.8 = 1/4 = 0.25.
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