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determine the quantity of atoms of vanadium in 1.28 grams of vanadium

Question

determine the quantity of atoms of vanadium in 1.28 grams of vanadium

Explanation:

Step1: Find molar mass of vanadium

The molar mass of vanadium (V) is approximately 50.94 g/mol.

Step2: Calculate number of moles

The number of moles $n$ of vanadium is calculated using the formula $n=\frac{m}{M}$, where $m = 1.28$ g and $M=50.94$ g/mol. So $n=\frac{1.28}{50.94}\text{ mol}$.

Step3: Use Avogadro's number

Avogadro's number $N_A = 6.022\times 10^{23}$ atoms/mol. The number of atoms $N$ of vanadium is $N=n\times N_A=\frac{1.28}{50.94}\times6.022\times 10^{23}$ atoms.
$N=\frac{1.28\times6.022\times 10^{23}}{50.94}\approx1.52\times 10^{22}$ atoms.

Answer:

$1.52\times 10^{22}$ atoms