Question

Naturally occurring boron consists of two isotopes whose atomic weights are 10.01 and 11.01. The atomic weight of natural boron is 10.81. Calculate the percentage of each isotope in natural boron.

Answer 1

Answer:

Let the occurrence of one form be x, then the occurrence of other form will be 100-x.

Putting this in the formula, we get,

[10.01*x + 11.01*(100-x)]/100 = 10.81

Solving this equation is not very hard, we can easily solve this equation as it consists only of a single variable.

10.01x - 11.01 x + 1101 = 1081

-x = -20

x = 20.

Therefore the occurrence of two isotopes of Boron is 20% and 80 %

Answer 2

Answer:

The percentage of each isotope in natural boron is 21.97 % and 78.03%.

Explanation:

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of first isotope be 'x'. So, fractional abundance of second  isotope will be '1 - x'

For first isotope:

Mass of first isotope = 10.01 amu

Fractional abundance of first isotope= x

For second isotope:

Mass of second isotope = 11.01 amu

Fractional abundance of second isotope= (1-x)

Average atomic mass of boron = 10.81 amu

Putting values in equation 1, we get:

$10.81=[(10.01 amu \times x)+(11.01 amu\times (1-x))$

x = 0.2197

Percentage abundance of first isotope = $0.2197\times 100=21.97\%$

Percentage abundance of second isotope = $(1-0.2197)=0.7803\times 100=78.03\%$