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Question: A metallic elements exists as a cubic lattice. Each edge of the unit cell is 2.88*Å*. The density of...

A metallic elements exists as a cubic lattice. Each edge of the unit cell is 2.88Å. The density of the metal is 7.20 gcm−3.gcm^{- 3}. How many unit cells will be present in 100 gm of the metal.

A

5.82 ×1023\times 10^{23}

B

6.33×10236.33 \times 10^{23}

C

7.49×10247.49 \times 10^{24}

D

6.9×10246.9 \times 10^{24}

Answer

5.82 ×1023\times 10^{23}

Explanation

Solution

The volume of unit cell (V) = a3 = (2.88Å)3

= 23.9×10−24cm323.9 \times 10^{- 24}cm^{3}

Volume of 100 g of the metal = MassDensity=1007.20=13.9cm2\frac{\text{Mass}}{\text{Density}} = \frac{100}{7.20} = 13.9cm^{2}

Number of unit cells in this volume = 13.9cm323.9×10−24cm3=5.82×1023\frac{13.9cm^{3}}{23.9 \times 10^{- 24}cm^{3}} = 5.82 \times 10^{23}