Question

If the atomic mass units of \(^{85}\text{Rb}\) and \(^{87}\text{Rb}\) are \(84.912\) and \(86.901\), respectively, then the abundance of \(^{87}\text{Rb}\) is _________ % (rounded off to nearest integer).
[Use: Average atomic mass of Rb = \(85.467\) amu]

Hint:

Set up a weighted average of the two isotope masses and solve for the unknown fraction.

Answer 1

Correct Answer: 28

Step 1: Set up the two isotope mixture.
Rubidium in nature is a mix of two isotopes, \(^{85}\text{Rb}\) with atomic mass \(84.912\) amu and \(^{87}\text{Rb}\) with atomic mass \(86.901\) amu. Let \(x\) be the fraction (as a decimal) of \(^{87}\text{Rb}\), so \((1-x)\) is the fraction of \(^{85}\text{Rb}\).

Step 2: Write the weighted average equation.
The average atomic mass is the mass-weighted sum of the two isotopes:
\[ 85.467 = 84.912(1-x) + 86.901x \]

Step 3: Solve for x.
Expand the right side:
\[ 85.467 = 84.912 - 84.912x + 86.901x \]
\[ 85.467 - 84.912 = (86.901 - 84.912)x \]
\[ 0.555 = 1.989x \]
\[ x = \frac{0.555}{1.989} = 0.2790 \]

Step 4: Convert to percentage and round.
The abundance of \(^{87}\text{Rb}\) is \(0.2790 \times 100 = 27.9\%\), which rounds to \(28\%\). This falls inside the accepted range of \(27\) to \(29\).

Final Answer:
\[ \boxed{28\%} \]