Question

Boron has two stable isotopes, 10B (relative abundance = 19%) and 11B (relative abundance = 81%). The atomic mass (in amu) that should appear for boron in the periodic table is :

• A. 10.8
• B. 10.2
• C. 11.2
• D. 10.6

Answer 1

Answer: (A)

10.8

Answer 2

(1)

Atomic mass

$= \frac{10 \times 19 \times 81 \times 11}{100} = \frac{190 + 1081}{100} = \frac{1081}{100} = 10.81$