Final answer:
To find the quantity of vanadium atoms in 1.28 grams, calculate the number of moles and then multiply by Avogadro's number resulting in approximately 1.51 × 1022 vanadium atoms.
Step-by-step explanation:
To calculate the quantity of vanadium atoms in 1.28 grams, we'll use the concept of moles and Avogadro's number. Firstly, we find the number of moles of vanadium in 1.28 grams by dividing the mass by the molar mass of vanadium (which is approximately 50.9415 g/mol). Then, we multiply this number by Avogadro's number (6.022 × 1023 atoms/mol) to find the total number of atoms.
- Determine the molar mass of vanadium from the periodic table: 50.9415 g/mol.
- Calculate the number of moles of vanadium in 1.28 grams: (1.28 g) / (50.9415 g/mol) = 0.02511 mol.
- Calculate the number of atoms by multiplying the moles by Avogadro's number: (0.02511 mol) × (6.022 × 1023 atoms/mol) = 1.51 × 1022 atoms.
Therefore, there are approximately 1.51 × 1022 atoms of vanadium in 1.28 grams.