The relative atomic mass of boron (which exists in two isotopic forms ^ 10 B and ^ 11 B ) is 10.81. What will be the abundance of ^ 10 B and ^ 11 B , respectively (consider a sample of 100 atoms)?

Correct answer: A. 19 \% and 81 \% Formula used, Average atomic mass = \% abundance × atomic mass total abundance aligned Let, \% of B^ 10 & =x \% of B^ 11 & =100-x average atomic mass & =10.81 10.81 & =(x × 10+(100-x) × 11)/(100) 1081 & =10 x+1100-11 x -19 & =-x x & =19 \% of B^ 10 & =19 \% \% of B^ 11 & =81 \% aligned

NDA 2024

Exam: NDA
Year: 2024
Subject: General Ability
Chapter: General Science

NDA 2024 · General Ability · General Science

The relative atomic mass of boron (which exists in two isotopic...

The relative atomic mass of boron (which exists in two isotopic forms

{ }^{10} \mathrm{ B}

and

{ }^{11} \mathrm{ B}

) is 10.81. What will be the abundance of

{ }^{10} \mathrm{ B}

and

{ }^{11} \mathrm{ B}

, respectively (consider a sample of 100 atoms)?

Options

A $19 \%$ and $81 \%$
B $81 \%$ and $19 \%$
C $38 \%$ and $62 \%$
D $62 \%$ and $38 \%$

Correct answer

A. $19 \%$ and $81 \%$

Step-by-step solution

Formula used, Average atomic mass

=\frac{\Sigma \% \text{ abundance } \times \text{ atomic mass }}{\Sigma \text{ total abundance }}

\begin{aligned} \text{ Let, } \quad \% \text{ of } \mathrm{B}^{10} & =x \\ \% \text{ of } \mathrm{B}^{11} & =100-x \\ \text{ average atomic mass } & =10.81 \\ 10.81 & =\frac{x \times 10+(100-x) \times 11}{100} \\ 1081 & =10 x+1100-11 x \\ -19 & =-x \\ x & =19 \\ \% \text{ of } \mathrm{B}^{10} & =19 \% \\ \% \text{ of } \mathrm{B}^{11} & =81 \% \end{aligned}

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