# How many atoms of vanadium are present in "185 mg" of this element?

The answer is ${\text{2.19" * "10}}^{21}$ atoms of vanadium are in $\text{185 mg}$.
$\text{185 mg" * ("1 gram")/("1000 mg") * ("1 atom")/(8.46 * 10^(-23)"g") = 2.19 * 10^(21)"atoms}$
You could double check your answer by comparing the number of atoms in $\text{185 mg}$ to the number of atoms in 1 mole of vanadium, or $\text{50.9 g}$.
$\text{185 mg}$ represents a mass about 300, or $3 \cdot {10}^{2}$, times smaller than $\text{50.9 g}$, so that's how many times less atoms you'll have compared to the number of atoms in 1 mole, $6.022 \cdot {10}^{23}$.