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Answer :
Final answer:
We first calculate the moles of Lithium in 97.9 g (about 14.11 mol), then use Avogadro's number to translate moles to atoms, giving us roughly 8.49 × 10⁴⁴ Li atoms, matching option C.
Explanation:
The question you're asking is how many Lithium (Li) atoms are contained in 97.9 g of Li. To solve this, we'll first need to calculate the amount of moles in 97.9 g of Lithium. Lithium has a molar mass of approximately 6.94 g/mol. By dividing your given mass by the molar mass, we will find the mole quantity: 97.9 g / 6.94 g/mol ≈ 14.11 mol. Now we translate the quantity of moles into atoms using Avogadro's number (roughly 6.022 × 10²³ atoms/mole). Multiplying our moles by Avogadro's number we get: 14.11 mol x 6.022 × 10²³ atoms/mole ≈ 8.49 × 10²⁴ Li atoms. Therefore, the number of Lithium atoms in 97.9 g of Lithium is approximately 8.49 × 10²⁴, matching option C).
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