The number of atoms in 1.28 grams of vanadium is 1.51×10²² atoms
Avogadro's hypothesis
From Avogadro's hypothesis,
1 mole of Vanadium = 6.02×10²³ atoms
But
1 mole of Vanadium = 51 g
Thus,
51 g of Vanadium = 6.02×10²³ atoms
How to determine the atoms in 1.28 g of Vanadium
51 g of Vanadium = 6.02×10²³ atoms
Therefore,
1.28 g of Vanadium = (1.28 × 6.02×10²³) / 51
1.28 g of Vanadium = 1.51×10²² atoms
Thus, 1.51×10²² atoms is present in 1.28 g of Vanadium
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