Chemistry 2011 JAMB Past Questions - Question 7: relative atomic mass of a naturally occurring consisting of 90% Li and 10% Li is

Correct Answer

A

Explanation

To calculate the relative atomic mass of a naturally occurring sample of lithium consisting of 90% lithium-7 (\(^7\)Li) and 10% lithium-6 (\(^6\)Li), you can use the weighted average formula:

\[ \text{Relative atomic mass} = ( \text{isotopic mass} \times \text{fractional abundance}) + ( \text{isotopic mass} \times \text{fractional abundance}) \]

For lithium:
- \( \text{Isotopic mass of } ^6\text{Li} = 6.015 \)
- \( \text{Isotopic mass of } ^7\text{Li} = 7.016 \)
- Fractional abundance of \( ^6\text{Li} = 0.10 \)
- Fractional abundance of \( ^7\text{Li} = 0.90 \)

Now, plug in the values:

\[ \text{Relative atomic mass} = (6.015 \times 0.10) + (7.016 \times 0.90) \]
\[ \text{Relative atomic mass} = 0.6015 + 6.3144 \]
\[ \text{Relative atomic mass} = 6.9159 \]

Rounded to the nearest whole number, the relative atomic mass of this sample is approximately 6.9.

So, the correct answer is: \( \boxed{6.9} \)