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Question

Boron found in nature has an atomic weight of 10.811 and is made up of the isotopes B10 (mass 10.013 amu) and B11 (mass 11.0093). What percentage of naturally occurring boron is made up of B10 and B11, respectively?

A
30 : 70
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B
25 : 75
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C
20 : 80
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D
15 : 85
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Solution

The correct option is C 20 : 80
Let the abundance of B11 be x% and B10 be (100x)%
Average atomic mass= [Atomic mass of B11× abundance + Atomic mass of B10× abundance]/100
10.811=11.0093×x(percent)+10.013×(100x)(percent)100
10.811×100=11.0093x%+1001.310.013x%
1081.1=0.9963x%+1001.3
1081.11001.3=0.9963x%
79.8=0.9963x%
x%=79.80.9963
x%=80
(100x)%=20
Natural abundance of B10=20
Natural abundance of B11=80
Ratio= 20:80

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