the density of alumnium is 2.7 g/cm^3 and its molar mass is 27 g/mol. find the mass of an atom of aluminum and the numebr of alumuminum atoms per cubic meter.
2 answers
Find the volume of one mole (27/2.7 cm^3). Divide that by avag number, that will be the volume per atom.
Molar mass is the mass in grams of 1 mole of Al, that is, 1 mol Al has a mass of 27 g.
On the other hand, 1 mole of any element contains Avogadro's number of atoms, that is,
1 mol Al = 6.02x10^23 atoms of Al
Therefore; 6.02x10^23 atoms of Al = 27 g
1 single Al atom;
27 g / 6.02x10^23 atoms of Al = 4.48x10^-23 g
d = 2.7 g / cm^3 means that 1 cm^3 Al has a mass of 2.7 g.
1 m^3 = 10^6 cm^3 = 2.7x10^6 g
Mole of Al = 2.7x10^6 g / 27 g/mol = 1x10^5 mol Al
Since each mole contains 6.02x10^23 atoms of Al
1x10^5 mol Al will contain;
1x10^5 mol Al x 6.02x10^23 atoms of Al /mol =
6.02x10^28 atoms of Al will be per m^3 of Al
On the other hand, 1 mole of any element contains Avogadro's number of atoms, that is,
1 mol Al = 6.02x10^23 atoms of Al
Therefore; 6.02x10^23 atoms of Al = 27 g
1 single Al atom;
27 g / 6.02x10^23 atoms of Al = 4.48x10^-23 g
d = 2.7 g / cm^3 means that 1 cm^3 Al has a mass of 2.7 g.
1 m^3 = 10^6 cm^3 = 2.7x10^6 g
Mole of Al = 2.7x10^6 g / 27 g/mol = 1x10^5 mol Al
Since each mole contains 6.02x10^23 atoms of Al
1x10^5 mol Al will contain;
1x10^5 mol Al x 6.02x10^23 atoms of Al /mol =
6.02x10^28 atoms of Al will be per m^3 of Al
4 answers