Modify Radius of Iron for Density Accuracy

[Ryo124]

Homework Statement

The atomic mass of iron (Fe) is approximately 56 grams per mole.

(a) Assume each iron atom in a bar of iron takes up a volume equal to a sphere of radius 2.35×10-10 m. Calculate the density of iron based on this model.

(b) The actual density of iron is 7870 kg/m^3. Modify the atomic radius in (a) such that the calculated density agrees with the actual value.

Homework Equations

Density = mass/(4/3)$$\pi$$r^3 ( D=m/V )

The Attempt at a Solution

I figured out a) to be 1.71E3 kg/m^3. However, I can't figure out b).

The new radius is 1.41E-9 from what I calculated , but it is not the right answer.

 

[Ryo124]

Does anyone know? Can anyone help?

 

[azatkgz]

You just need to find how many atoms are in one mole.

One mole weighs 56 grams,and density is

$$\rho=\frac{56 g}{Nv_0}$$, where $$v_o$$ is a volume of one atom.

 

[Ryo124]

Thanks azat for making me realize my mistake. I was calculating the actual mass to be in g and not kg (I had already converted N) while leaving the density in kg/m^3. Thank you, it turns out to be 1.41E-10 m.