Question

a sample of chlorine contains the following relative abundances of its isotopes.
isotope | relative abundance
$ce{_{17}^{35}cl}$ | 75.78%
$ce{_{17}^{37}cl}$ | 24.22%
what is the average relative atomic mass of the chlorine in the sample?

Explanation:

Step1: Convert percentages to decimals

For \(^{35}_{17}\text{Cl}\), \(75.78\% = 0.7578\); for \(^{37}_{17}\text{Cl}\), \(24.22\% = 0.2422\).

Step2: Calculate contribution of each isotope

Contribution of \(^{35}_{17}\text{Cl}\): \(35\times0.7578 = 26.523\)
Contribution of \(^{37}_{17}\text{Cl}\): \(37\times0.2422 = 8.9614\)

Step3: Sum the contributions

Total atomic mass: \(26.523 + 8.9614 = 35.4844\)

Answer:

\(35.48\) (or \(35.4844\) depending on precision)