Question

the relative atomic mass of boron is 10.8. calculate the % of isotopes 10B and 11B occuring in nature

Answer 1

Weighted average:

(x/100)*10 + [(100-x)/100]*11 = 10.8 where x is the percentage of 10B isotope

10x/100 + (1100-11x)/100 = 10.8

-x +1100 = 1080

x = 20

10B = 20%

11B = 80%

Answer 2

The percentage abundance of $_{5}^{10}\textrm{B}$ and $_{5}^{10}\textrm{B}$ isotopes are 20% and 80% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

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 $_{5}^{10}\textrm{B}$ isotope be 'x'. So, fractional abundance of $_{5}^{11}\textrm{B}$ isotope will be '1 - x'

For $_{5}^{10}\textrm{B}$ isotope:

Mass of $_{5}^{10}\textrm{B}$ isotope = 10 amu

Fractional abundance of $_{5}^{10}\textrm{B}$ isotope = x

For $_{5}^{11}\textrm{B}$ isotope:

Mass of $_{5}^{11}\textrm{B}$ isotope = 11 amu

Fractional abundance of $_{5}^{11}\textrm{B}$ isotope = 1 - x

Average atomic mass of boron = 10.8 amu

Putting values in equation 1, we get:

$10.8=[(10\times x)+(11\times (1-x))]\\\\x=0.2$

Percentage abundance of $_{5}^{10}\textrm{B}$ isotope = $0.2\times 100=20\%$

Percentage abundance of $_{5}^{11}\textrm{B}$ isotope = $(1-0.2)=0.8\times 100=80\%$

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