Boron has two naturally occurring isotopes, 10-B (isotopicmass 10.0129 amu) and

Question

Boron has two naturally occurring isotopes, 10-B (isotopicmass 10.0129 amu) and 11-B (isotopic mass = 11.0093 amu). Given the atomic mass of boron (10.81 amu), determine the percentabundance of boron-11........... I got 7.4%, but my professor's answer is 80%, HOW????thanks Boron has two naturally occurring isotopes, 10-B (isotopicmass 10.0129 amu) and 11-B (isotopic mass = 11.0093 amu). Given the atomic mass of boron (10.81 amu), determine the percentabundance of boron-11........... I got 7.4%, but my professor's answer is 80%, HOW????thanks

Explanation / Answer

Let X be the percent abundance of 10-B and Y be the percentabundance of 11-B 10.81 = (11.0093X + 10.0129Y) / (X+ Y) 10.81X + 10.81Y = 11.0093X + 10.0129Y 0.7971Y = 0.1993X X = 4Y X + Y = 100 4Y + Y = 100 5Y = 100 Y = 20 X = 4Y = 4(20) = 80 Thus, the percent abundance of boron-11 is 80% Hope this helps!

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