Final answer:
There are approximately 3.421 × 10^21 atoms of bromine in 49 milligrams of PBr₅. This is calculated by first determining the moles of PBr₅ from the given mass and then multiplying by Avogadro's number and the number of bromine atoms per molecule of PBr₅.
Step-by-step explanation:
To determine how many atoms of bromine are in 49 milligrams of PBr₅, we need to perform a few calculations. Phosphorus pentabromide (PBr₅) has a molecular weight of (30.97 for P) + (5 × 79.904 for Br) = 430.87 g/mol. In order to find the moles of PBr₅ in 49 milligrams (0.049 grams), we use the formula: moles = mass (g) / molar mass (g/mol).
Moles of PBr₅ = 0.049 g / 430.87 g/mol = 1.137 × 10^-4 mol.
Since each molecule of PBr₅ contains 5 bromine atoms, we multiply the moles of PBr₅ by Avogadro's number (6.022 × 10^23 atoms/mol) to get the total number of bromine atoms:
Bromine atoms = 1.137 × 10^-4 mol × 5 × 6.022 × 10^23 atoms/mol = 3.421 × 10^21 atoms of bromine.
Therefore, there are approximately 3.421 × 10^21 atoms of bromine in 49 milligrams of PBr₅.