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Answer :
Final answer:
The number of atoms in 37.1 grams of LiBr is approximately 2.574 × 10²³ atoms.
Explanation:
To calculate the number of atoms in 37.1 grams of LiBr, we need to start by determining the molar mass of LiBr. Lithium (Li) has a molar mass of 6.941 grams/mol, and Bromine (Br) has a molar mass of 79.904 grams/mol. Adding these two values together gives us the molar mass of LiBr:
6.941 grams/mol + 79.904 grams/mol = 86.845 grams/mol.
Next, we can use Avogadro's number (6.022 × 10²³ mol⁻¹) to convert the given mass of LiBr to moles. To do this, we divide the given mass (37.1 grams) by the molar mass (86.845 grams/mol).
37.1 grams / 86.845 grams/mol = 0.427 moles of LiBr.
Finally, we can use Avogadro's number again to convert the number of moles to the number of atoms. Multiply the number of moles (0.427) by Avogadro's number (6.022 × 10²³ mol⁻¹) to get approximately 2.574 × 10²³ atoms of LiBr.
0.427 moles * 6.022 × 10²³ mol⁻¹ = 2.574 × 10²³ atoms
Learn more about Avogadro's number here:
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