Question

Boron has two isotopes with atomic masses 10 and 11. If its average atomic mass is 10.81, the abundance of the lighter isotope is:

• A. 19\%
• B. 20\%
• C. 25\%
• D. 10\%

Hint:

For isotopic abundance problems, use the formula: \[ \text{Average mass} = \frac{\text{(Mass of isotope 1) \(\times\) (abundance)} + \text{(Mass of isotope 2) \(\times\) (abundance)}}{100}. \] Simplify step-by-step for accurate results.

Answer 1

The Correct Option is A

Let the percentage abundance of the lighter isotope (\( \mathrm{^{10}B} \)) be \( x \%\). Then the percentage abundance of the heavier isotope (\( \mathrm{^{11}B} \)) will be \( (100 - x)\%\). The average atomic mass of boron is given by: \[ \text{Average atomic mass} = \frac{(x \cdot 10) + ((100 - x) \cdot 11)}{100}. \] Substitute the given average atomic mass \( 10.81 \): \[ 10.81 = \frac{(x \cdot 10) + ((100 - x) \cdot 11)}{100}. \] Simplify: \[ 10.81 = \frac{10x + 1100 - 11x}{100}. \] Combine like terms: \[ 10.81 = \frac{1100 - x}{100}. \] Multiply through by 100: \[ 1081 = 1100 - x. \] Solve for \( x \): \[ x = 1100 - 1081 = 19. \] Thus, the abundance of the lighter isotope is \( \mathbf{19\%} \).