Modify Radius of Iron for Density Accuracy

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The discussion revolves around calculating the density of iron using a model where each atom is represented as a sphere with a specified radius. Initially, the calculated density was found to be 1.71E3 kg/m^3, which is significantly lower than the actual density of iron at 7870 kg/m^3. The correct atomic radius needed to match the actual density was determined to be 1.41E-10 m, after correcting a mistake in unit conversion from grams to kilograms. The participants emphasized the importance of accurately determining the number of atoms in a mole to achieve the correct density calculation. Ultimately, the thread highlights the significance of unit consistency in density calculations.
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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)\pir^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.
 
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Does anyone know? Can anyone help?
 
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.
 
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.
 
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