Density of aluminum foil: 2.70 g/ml

atomic mass: 26.98g/mol

radius of aluminum atom: 143 picometer

packing density: 74%

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.

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