How many unit cells are present in a cubeshaped ideal crystal of NaCl of mass 1.00 g.? (A) 5.14*1021 unit cells(B) 1.28*1021 unit cells(C) 1.71*1021 unit cells(D) 2.57*1021 unit cells

we can see that:

• Na+ ions are present on all the twelve edge centres and also in centre of the cube.
• Cl- ions are present on all the eight corners as well as all the six face centres.

Here are some points to note before proceeding to calculations:

1. As the atoms present on edges contribute 1/4th of their volume to each surrounding unit cells.
2. The atom present at centre of the unit cell contribute full one atom
3. The atoms present at face centre contribute half of it's volume to each surrounding unit cells.
4. The atoms present at the corners contribute 1/8th of their volume to each surrounding unit cells.

Let's calculate:

The number of Na+ ions present in one unit cell = 12×1/4 + 1 =4 atoms per unit cell.

The number of Cl- ions present in one unit cell = 6×1/2 + 8×1/8 = 4 atoms per unit cell.

So, each unit cell has 4 atoms of each Na+ and Cl-. Hence there are total 4 molecules of NaCl in each unit cell.

As the mass of -Avogadro number of molecules- of NaCl is 58.5g.

Then, number of molecules in one gram = 6.022× 10^23 ÷ 58.5 = 1.03×10^22 molecules

As we found above that each unit cell contains 4 molecules of NaCl hence we can now easily find the number of unit cells containing 1.03×10^22 molecules = 1.03×10^22 ÷ 4 = 2.575×10^21 unit cells

So, 2.575×10^21 unit cells contains 1gm of NaCl molecules.