Isotopes — practice questions

Step 1: Apply the formula for relative atomic mass. $A_r = \frac{(\text{mass}_1 \times \text{\%}_1) + (\text{mass}_2 \times \text{\%}_2)}{100}$

Step 2: Substitute the given values into the formula. $A_r = \frac{(34.97 \times 75.77) + (36.97 \times 24.23)}{100}$

Step 3: Calculate the contribution from each isotope. Contribution from $^{35}\text{Cl}$: $34.97 \times 75.77 = 2649.5769$ Contribution from $^{37}\text{Cl}$: $36.97 \times 24.23 = 895.7831$

Step 4: Sum the contributions and divide by 100. $A_r = \frac{2649.5769 + 895.7831}{100} = \frac{3545.36}{100}$

Step 5: State the final answer to the required precision. $A_r = 35.4536$ To two decimal places, $A_r = 35.45$

[1 mark for correct formula/substitution, 1 mark for the final correct answer]

Step 1: Define variables for the unknown abundances. Let the percentage abundance of $^{10}\text{B}$ be $x$. Since there are only two isotopes, the percentage abundance of $^{11}\text{B}$ must be $(100 - x)$.

Step 2: Set up the equation for relative atomic mass. $10.81 = \frac{(10.01 \times x) + (11.01 \times (100 - x))}{100}$

Step 3: Solve the equation for $x$. First, multiply both sides by 100. $1081 = 10.01x + 1101 - 11.01x$

Step 4: Rearrange the equation to isolate terms with $x$. $1081 = 1101 - 1.00x$ $1.00x = 1101 - 1081$ $1.00x = 20$ $x = 20$

Step 5: State the final answer clearly. The percentage abundance of $^{10}\text{B}$ is 20%. (The abundance of $^{11}\text{B}$ is therefore 80%).

[1 mark for setting up the abundance as x and 100-x, 1 mark for the correct algebraic setup, 1 mark for the final correct answer]