Chlorine consisting of two isotope of mass number 35 and 37 in the ratio 3:1 has an atomic mass of 35.5…

Analysis

The question asks for the relative abundance (percentage) of the chlorine isotope with a mass number of 37, given the ratio of the two isotopes ($^{35}\text{Cl}$ and $^{37}\text{Cl}$) is 3:1.

Concept

Relative atomic mass is the weighted average of the masses of all naturally occurring isotopes of an element. The formula is:

Where the sum of the relative abundances must equal 100%.

Solution

1. Understand the Ratio: The isotopes are in the ratio 3:1. This means for every 4 atoms of chlorine, 3 are $^{35}\text{Cl}$ and 1 is $^{37}\text{Cl}$.
2. Calculate the Total Parts: The total number of parts in the ratio is $3 + 1 = 4$.
3. Calculate the Fraction of $^{37}\text{Cl}$: The isotope with mass number 37 represents 1 part out of 4. $$\text{Fraction} = \frac{1}{4}$$
4. Convert to Percentage: To find the relative abundance (percentage), multiply the fraction by 100: $$\text{Relative Abundance} = \left( \frac{1}{4} \right) \times 100\% = 25\%$$

(Self-check: The relative abundance of $^{35}\text{Cl}$ would be $\frac{3}{4} \times 100\% = 75\%$. Calculating the average mass: $(35 \times 0.75) + (37 \times 0.25) = 26.25 + 9.25 = 35.5$. This confirms our calculation is correct.)

Conclusion

The relative abundance of the isotope with mass number 37 is 25%. This matches option B.

Exam Tip

When you see a ratio like $x:y$, the percentage of the second component is always $\frac{y}{x+y} \times 100$. Don't waste time doing complex algebraic equations if the ratio is already provided; simply convert the ratio parts into a percentage of the total!