Question

Boron has two isotopes B-10 and B-11. The average atomic mass of boron is found to be 10.80u .calculate the percentage of abundance of these isotopes

Answer 1

m1p1+m2p2/100 = avg atomic mass
take p1=x and p2=100-x and hence find x by using abive formula

Answer 2

$\huge\fbox\blue{Question}$

Boron has two isotopes B -10 and B -11. The average atomic mass of boron is found to be 10.804 amu. Calculate the percentage of abundance of these isotopes

$\huge\mathfrak\red{Answer:- \:80.4\%}$

$\huge\fbox\blue{Solution}$

Let % Relative abundance of B-10 =x

Let % Relative abundance of B-11 = (100 - x)

Average atomic mass of Boron:

$= \frac{Mass \: of \: 1st \: Isotope \times \: Abundance \: of \: 1st \: Isotope \: \times \: Mass \: of \: 2nd \: Isotope \: \times \: Abundance \: of \: 2nd \: Isotope }{100}$

$10.804 = \frac{10x + 11 \: (100 - x)}{100}$

$10.804 = \frac{10x \: + \: 1100 - 11 x)}{100}$

⇒ 10.804 × 100 = 1100 - x

⇒ 1080.4 = 1100 - x

⇒ x = 1100 - 1080.4

⇒ x = 196%

% Relative abundance of B-10

⇒x=19.6%

% Relative abundance of B-11 ⇒100-x

⇒ 100 - 9.6

⇒ y = 80.4 %

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