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Question

Chemistry Question on The solid state

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

A

6.85×1026.85 \times 10^2

B

5.82×10235.82 \times 10^{23}

C

4.37×1054.37 \times 10^5

D

2.12×1062.12 \times 10^6

Answer

5.82×10235.82 \times 10^{23}

Explanation

Solution

The volume of the unit cell =(2.88 A˚)3=23.9×10−24 cm3= (2.88 \, \mathring A )3 = 23.9 \times 10^{-24} \, cm^3. The volume of 100 g of the metal =mρ=1007.20=13.9 cm3 = \frac{m}{\rho} = \frac{100}{7.20} = 13.9 \, cm^3 Number of unit cells in this volume =13.9 cm323.9×10−24 cm3=5.82×1023 = \frac{13.9 \, cm^3}{23.9 \times 10^{-24} \, cm^3} = 5.82 \times 10^{23}