How many unit cells are present in 2.5 gram of cubic crystal of kcl with each edge of cells
Answers
Here we can see that :
1) K+ ions are present on all the 12 edge centres and also in centre of cube.
2) Cl- ions are present on all the 8 centres as well as all the 6 face centres.
Before calculations, note that:
I) As the atoms present on edges contribute 1/4th of their volume to each surrounding unit cells.
II) The atom present at centre of the unit cell contribute full one atom.
III) The atoms present at face centre contribute half of its volume to each surrounding unit cells.
IV) The atoms present at the corners contribute 1/8th of their volume to each surrounding unit cells.
Calculations:
The number of K+ ions present in one unit cell = 12 *1/4 + 1 = 4 atoms / unit cell.
The number of Cl- ions present in one unit cell = 6*1/2 + 8*1/8 = 4 atoms/unit cell.
So, each unit cell has 4 atoms of each k+ and Cl - ions. Hence there are total 4 molecules of KCl in each unit cell.
As the mass of Avogadro number of molecules of KCl is 74.5 g. Then, number of molecules in 2.5 g =6.022*10^23/74.5
=0.08083*10^23
=8.083*10^21 molecules.
As we found above that each unit cell contains 4 molecules of KCl. Thus, the number of unit cells containing 8.083k10e21 molecules = 8.083*10^21/4
= 2.02075*10^21
So, 2.02075*10^21 unit cells contains 2.5 g of KCl molecules.