# Density and Number of Atoms

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The Density is 7.87g/cm^3 , and the average mass for a single iron atom is 9.28*10^-23g .

How many iron atoms are present in an iron bar whose dimensions are 2.00 cm * 3.00 cm * 8.00 cm ?

https://brainmass.com/chemistry/stoichiometry/density-number-atoms-7975

## SOLUTION This solution is **FREE** courtesy of BrainMass!

We have

Mass/Volume = Density

The volume of the iron bar is V = 2 * 3 * 8 cm^3

V = 48 cm^3

Since the density of iron is given to be 7.87g/cm^3 we can calculate the mass of the iron bar as,

mass = density * volume = 48 * 7.87 = 377.76 g

This is the total mass of all the iron atoms in the bar. But each iron atom has a mass, m = 9.28 * 10^-23g

We have 377.76g as the mass of N atoms in the bar

ie, N*m = 377.76

Now, N = 377.76/9.28 * 10^-23 = 40.7*10^23

The number of iron atoms present in the bar is 40.7*10^23 (ans)

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