Question

How many unit cells are present in a cube shaped ideal crystal of nacl of mass 1g?

Answer 1

here is the answer for you dear

Answer 2

$2.575\times10^2^1$ unit cells present in a cube shaped ideal crystal of nacl of mass 1g.

Explanation:

• The number of$Na^+$ ions existing in one unit cell =$12\times \frac 14 + 1 =4$ atoms per unit cell.

• The number of$Cl^-$ ions existing in one unit cell =$6\times \frac12 + 8\times \frac18 = 4$ atoms per unit cell.

• Every individual unit cell has 4 atoms of each$Na^+$ and $Cl^-$.

• Hence there are a "total of 4 molecules of NaCl  which is present in each unit cell.

• The mass of Avogadro’s number of molecules of NaCl is 58.5g.

• Hence number of molecules in one gram = $\frac{6.022 \times 10^{23}}{58.5}=1.03 \times 10^{22}$ molecules

• Now, each unit cell contains 4 molecules of NaCl

• Hence the number of unit cells containing $1.03\times10^2^2$ molecules =$\frac{1.03 \times 10^{22}}{4}=2.575 \times 10^{21}$ unit cells.

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