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## Homework Statement

Two questions :

1. find the (mass) density of Sodium. Atomic weight: 22.9898. Crystal structure: bcc. a=4.23 angstrom (crystal constant).

2. Given the (mass) density of Aluminum : 2.7 g/cm^3. Atomic weight: 26.982. Find the number of atoms density (# of atoms per [itex]cm^{3}[/itex]).

## Homework Equations

These questions looks trivial, but my answers are slightly different from the given solutions.

I think that this might be because I have a mistake in the conversion from atomic weight to the weight of an atom.

To my understanding, the atomic weight is the weight in grams of Avogadro number ([itex]6.022\times10^{23}[/itex]) of atoms.

## The Attempt at a Solution

Therefore I get:

1. The weight of an atom is : [itex]\frac{22.9898}{6.022\times10^{23}}=3.818\times10^{-23} gram[/itex]. And since in bcc there are 2 atoms in unit cell we have density of:

[itex]\frac{2\times3.818\times10^{-23}}{{(4.23\times10^{-8})}^{3}}=1.009 g/cm^{3}[/itex]

But the given solution is [itex]0.97 g/cm^{3}[/itex]

2. The number of atoms density (# of atoms per [itex]cm^{3}[/itex]), is density/atom's weight. Atom's weight = [itex]\frac{26.982}{6.022\times10^{23}}=4.481\times10^{-23} gram[/itex]. Therefore, the atomic density is : [itex]\frac{2.7}{4.481\times10^{-23}}=6.025\times10^{23} atoms/cm^{3}[/itex].

But the given solution is [itex]6.028 \times10^{23} atoms/cm^{3}[/itex].

Where is my mistake?

Thanks,

Naftali