[Fish200398]

my calculations was

4/(6,02 x 10^23) moles x 27 g/moles = mass per unit cubic cell = 17,94 x 10^{-23} g

lenght side of cubic for fcc = x = 2,82 x r = 0,41 nm
volume of 1 unit cubic cell = x^3 = 0,0689 10^{-21} cm^3 = 6,89 x 10^{-23} cm^3

density = mass/volume =  2,6 g/cm^3

why the key answer is 2,7 ... ..

[AWK]

Do not round coefficient 2·SQRT(2) or do it correctly

[Fish200398]

yes, but the key answer to radius is 0,41. while i previously using 0,403 nm. it said round it to second decimal. maybe in the answer 0,41.. in the calculation should be 0,403...

[Enthalpy]

With the data provided, and the same computation, I get 2711kg/m³.
410pm instead of 404.5pm make the difference between 2600 and 2700. The length is cubed.

[MNIO]

let me try it my usual way

       26.98g    1 mol Al atoms      4 atoms                                   1 cell
ρ = ---------- x ------------------ x ---------- x ------------------------------------------------
        1 mol      6.022e23 atoms       cell          ((4 * 0.143nm * (100cm / 10^9nm) / √2)³

  = 2.7 g/cm³

*******
do you understand this?  do you understand where the volume of the cell came from?  that's the  (4r / √2)³ term

[Fish200398]

@MNIO isnt the x lenght = r sqrt 8 ?

[MNIO]

first of all, let's make sure you understand how the atoms in a unit cell are arranged, 

These are the corner atoms


notice how the corner atoms are only partially in each cell?  now this is the face of a fcc cell


let's do some algebra
  V = a³        agreed?  this is a cube and "a" is the edge length
and
  a² + a² = (4r)²            thank you Pythagoras.
  2a² = (4r)²
  a² = (4r)² / 2
  a = (4r) / √2
subbing
  V = (4r / √2)³

[Fish200398]

4r/sqrt 2 = r sqrt 8 right?

But to be safe, just plug in r which is known into 4r/sqrt 2 and take the 3rd power?

[AWK]

Your algebra was correct but you rounded intermediate results that caused to much error, eq 2SQRT(2)=2.82 instead of 2.823 and so on