Question

A metallic element exists as cubic lattice. Each edge of the unit cell is $2.88 \, \mathring A$. The density of the metal is 7.20 g $cm^{-3}$. How many unit cell will be present in 100 g of the metal?

• A. $6.85 \times 10^2$
• B. $5.82 \times 10^{23}$
• C. $4.37 \times 10^5$
• D. $2.12 \times 10^6$

Answer 1

Answer: (B)

The volume of the unit cell $= (2.88 \, \mathring A )3 = 23.9 \times 10^{-24} \, cm^3$.
The volume of 100 g of the metal
$ = \frac{m}{\rho} = \frac{100}{7.20} = 13.9 \, cm^3$
Number of unit cells in this volume
$ = \frac{13.9 \, cm^3}{23.9 \times 10^{-24} \, cm^3} = 5.82 \times 10^{23}$