How do i compute the numerical value for avogadro's number using the physical data for aluminum

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Density of aluminum foil: 2.70 g/ml

atomic mass: 26.98g/mol

radius of aluminum atom: 143 picometer

packing density: 74%

Oct 22nd, 2014

We shall proceed like this

we shall calculate the volume of one atom of aluminium using the vol of sphere formula with the given radius of 143 picometre

4/3x22/7x143x143x143 x10^(-30) cubic centimetre=1.2254x10^(-23) cubic cm

inverse of this number is 0.82x10^23 atoms packed in 1 cubic cm

As the actual packing density is 74% the volume they occupy in the foil is 1/0.74=1.35 cubic centimetre

So the weight of these many atoms works out to be 1.35x2.7g/cubic centimetre=3.65 gm

So the packed volumeof atoms require for one atomic weight will be 

26.98/3.65 cubic centimetre

The number of atoms will work out in one atomic weight in gms will be 26.98/3.65 x0.82x10^23

this works out to be 6.06x10^23 which is the Avogadro number by this method of calculation

The error is due to approximation in calculating to the max of fourth decimal place decimal place.

Nov 7th, 2014

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