Answer:
339 g of Rb.
Step-by-step explanation:
First, we have to convert 2.39 x 10²⁴ atoms of rubidium (Rb) using Avogadro's number which is 6.022 x 10²³ atoms/mol. The conversion will look like this:
![2.39\cdot10^(24)\text{ atoms Rb}\cdot\frac{1\text{ mol}}{6.022\cdot10^(23)\text{ atoms}}=3.968\text{ moles Rb}\approx3.97\text{ moles Rb,}](https://img.qammunity.org/2023/formulas/chemistry/college/21xaboj0elr3gxom0rcm1rc49spvxb0yup.png)
and now, let's convert this to grams using the molar mass of Rb which is 85.5 g/mol:
![3.97\text{ moles Rb}\cdot\frac{85.5\text{ g Rb}}{1\text{ mol Rb}}=339.4\text{ g Rb}\approx339\text{ g Rb.}](https://img.qammunity.org/2023/formulas/chemistry/college/5ui8qbw756dgph602nd4qee0m889kzsmha.png)
With correct significant figures (3 significant figures), the answer would be 339 g of Rb.