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Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10

Calculate the % relative abundance of B-10 and B-11, if its average atomic mass is 10.804 amu. - Science

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Question

Calculate the % relative abundance of B-10 and B-11, if its average atomic mass is 10.804 amu.

Sum

Solution

The average atomic mass of Boron = 10.804 amu.

% relative abundance of B – 10 =?

% relative abundance of B – 11 =?

Let the fraction of relative abundance of B – 10 = x

Let the fraction of relative abundance of B – 11 = y

x + y = 1

y = 1 – x

Relative abundance = x (10) + (1 – x) (11) = 10.804 amu

⇒ 10x + 11 – 11x = 10.804 amu

⇒ 11 – x = 10.804 amu

⇒ -x = 10.804 – 11

⇒ -x = -0.196

⇒ x = 0.196

x = % abundance of B – 10 = 0.196 × 100 = 19.6 %

y = % abundance of B – 11 = 100 – 19.6 = 80.4 %

Percentage abundance of B – 10 = 19.6 %

Percentage abundance of B – 11 = 80.4 %.

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Chapter 7: Atoms and Molecules - Evaluation [Page 104]

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Samacheer Kalvi Science [English] Class 10 SSLC TN Board
Chapter 7 Atoms and Molecules
Evaluation | Q IX. 4. | Page 104
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