The natural abundance for boron isotopes is 19.9% 10B and 80.1% 11B Calculate boron's atomic mass.

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

read more: link

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

where's the diagram?