Question

Boron occurs in nature in the form of two isotopes 5B 11,5B 10 in the ratio of 81% and 19% respectively calculate its average atomic mass​

Answer 1

Answer ⇒ 540.5 amu.

Explanation ⇒

% of ⁵B₁₁ Isotope = 81%

% of ⁵B₁₀ Isotope = 19%

Using the formula,

Average atomic mass = (mass no. × % of ⁵B₁₁ + mass no. × ⁵B₁₀)/2

∴ Average atomic mass = (11 × 81 + 10 × 19)/2

= (891 + 190)/2

= 1081/2

= 540.5 amu.

Hence, the average atomic mass of Boron is 540.5 amu.

Hope it helps.

Answer 2

Answer:

10.81%

Explanation:

avg. mass percentage = (mass number * percent)/total percentage

                                     = (11*81 + 10*19)/100 = 10.81%