Question

Q-9

9. How many unit cell are present in a cube shaped ideal crystal of NaCl of mass 1.00 g?
[Atomic masses Na = 23, Cl = 35.5]
a. 2.57 $\text{×}$ 10²¹
b. 5.14 $\text{×}$ 10²¹
c. 1.28 $\text{×}$ 10²¹
d. 1.71 $\text{×}$ 10²¹

Answer 1

Dear Student,
density = $\frac{Z\times M}{a^3\times N_A}$
Density = $\frac{mass}{volume}$ =$\frac{m}{a^3}$
mass = density $\times$volume = 1 gm

Number of unit cells = $\frac{mass\times avogadro\text{'}snumber}{Molecularmass\times z}$
where 'Z'= 4
putting the values we get

$\frac{1\times 6.023\times {10}^{23}}{58.5\times 4}$= 2.57 $\times$10²¹ unit cells
Regards!