Answer:
10.801 amu
Explanation:
From the question given above, the following data were obtained:
Isotope A (¹⁰B):
Mass of A = 10
Abundance (A%) = 19.9%
Isotope B (¹¹B):
Mass of B = 11
Abundance (B%) = 80.1%
Atomic mass of Boron =?
The atomic mass of boron can be obtained as illustrated below:
Atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]
= [(10 × 19.9)/100] + [(11 × 80.1)/100]
= 1.99 + 8.811
= 10.801 amu
Thus, the atomic mass of boron is 10.801 amu
The atomic mass of boron with natural abundance of 19.9% of 10 B and 80.1% of 11 B is 10.801 amu
Boron has 2 isotopes.
First isotopes
mass = 10
% abundance = 19.9%
Second Isotopes
mass = 11
% abundance = 80.1%
Therefore,
Atomic mass = (19.9% of 10) + (80.1% of 11)
Atomic mass = (19.9 / 100 × 10) + (80.1 / 100 × 11)
Atomic mass = 199 / 100 + 881.1 / 100
Atomic mass = 1.99 + 8.811
Atomic mass = 10.801
Atomic mass = 10.801 amu
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