(ii) The atomic masses of \( { }^{10} \mathrm{~B} \) and \( { }^{11} \mathrm{~B} \) are 10.0129 amu and 11.0093 amu, respectively. Calculate the natural abundances of these two isotopes. The average atomic mass of B is 10.81 amu. Answer: \( { }^{10} \mathrm{~B}=20 \% ; 11 \mathrm{~B}=80 \% \)

Given: - Atomic mass of \( { }^{10} \mathrm{~B} \) = 10.0129 amu - Atomic mass of \( { }^{11} \mathrm{~B} \) = 11.0093 amu - Average atomic mass of B = 10.81 amu Let's denote the natural abundance of \( { }^{10} \mathrm{~B} \) as \( x \% \) and the natural abundance of \( { }^{11} \mathrm{~B} \) as \( (100 - x) \% \). The average atomic mass of B can be calculated using the formula: \[ \text{Average atomic mass} = \frac{{\text{Atomic mass of } { }^{10} \mathrm{~B} \times x \% + \text{Atomic mass of } { }^{11} \mathrm{~B} \times (100 - x) \%}}{100} \] Substitute the given values into the formula: \[ 10.81 = \frac{{10.0129 \times x + 11.0093 \times (100 - x)}}{100} \] Now, we can solve this equation to find the natural abundance of \( { }^{10} \mathrm{~B} \) and \( { }^{11} \mathrm{~B} \). Solve the equation by following steps: - step0: Solve for \(x\): \(10.81=\frac{\left(10.0129x+11.0093\left(100-x\right)\right)}{100}\) - step1: Remove the parentheses: \(10.81=\frac{10.0129x+11.0093\left(100-x\right)}{100}\) - step2: Simplify: \(10.81=-\frac{2491}{250000}x+\frac{110093}{10000}\) - step3: Swap the sides: \(-\frac{2491}{250000}x+\frac{110093}{10000}=10.81\) - step4: Move the constant to the right side: \(-\frac{2491}{250000}x=10.81-\frac{110093}{10000}\) - step5: Subtract the numbers: \(-\frac{2491}{250000}x=-\frac{1993}{10000}\) - step6: Change the signs: \(\frac{2491}{250000}x=\frac{1993}{10000}\) - step7: Multiply by the reciprocal: \(\frac{2491}{250000}x\times \frac{250000}{2491}=\frac{1993}{10000}\times \frac{250000}{2491}\) - step8: Multiply: \(x=\frac{49825}{2491}\) The natural abundance of \( { }^{10} \mathrm{~B} \) is \( \frac{49825}{2491} \% \) or approximately 20%. Therefore, the natural abundance of \( { }^{10} \mathrm{~B} \) is 20% and the natural abundance of \( { }^{11} \mathrm{~B} \) is 80%.