Naturally occurring boron is 80.20% boron-11 (of atomic mass 11.01 amu) and 19.80% of some other isotope form of boron. What must be the atomic mass of this second isotope in order to account for the 10.81 amu average atomic mass of boron?

A. 10.99 amu
B. 11.00 amu
C. None of these
D. 11.01 amu
E. 11.99 amu
F. 12.00 amu
G. 10.00 amu
H. 9.00 amu

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  1. 0.8020(11.01) + 0.1980(x) = 10.81
    Solve for x.

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  2. X= 5.9089

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  3. Ans is 10.97amu

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