Question

Naturally occuring 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

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The percentage of isotope with A. wt 10.01 is 20%.

The percentage of isotope with A. wt 11.01 is 80%.

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Explaination :-

Let the percentage of isotope with atomic weight 10.01 be 'x'

Let the percentage of isotope with atomic weight 11.01 be '(100 - x)'

$\sf average \: atomic \: weight = \frac{ m_{1}x_{1} + m_{2}x_{2} }{x_{1} + x_{2} }$

$\sf 10.81 = \frac{x + 10.01 + (100 - x)11}{100}$

by calculating we get,

x = 20%

Thus,The percentage of isotope with A. wt 10.01 is 20%

and The percentage of isotope with A. wt 11.01 is (100 - x ) = 100 - 20 = 80%

Answer 2

Atomic weight of boron = 10.81

% of lighter isotope ( 10.01 B ) = ? (x)

% of heavier isotope ( 11.01 B ) = ( 100 - x ) ( 80% )

Atomic weight = Σ x of abundance × Mass number / 100

10.81 = ( x × 10.01 ) + ( 100 - x ) 11.01 / 100

= 10.81 × 100 = 10.01 + ( 100 × 11.01 ) - ( x × 11.01 (

1091 = 10.01 x + 11.02 - 11.01x

= 1081 - 1101 = 10.01 x - 11.01 x

= 20