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?

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 × 102 (B) 5.82 × 1023 (C) 4.37 × 105 (D) 2.12 × 106

Tardigrade Admin · Accepted Answer

The volume of the unit cell = (2.88 mathring A )3 = 23.9 × 10-24 cm3. The volume of 100 g of the metal = (m/ρ) = (100/7.20) = 13.9 cm3 Number of unit cells in this volume = (13.9 cm3/23.9 × 10-24 cm3) = 5.82 × 1023

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

$6.85 \times 1 0^{2}$

$5.82 \times 1 0^{23}$

$4.37 \times 1 0^{5}$

$2.12 \times 1 0^{6}$

The volume of the unit cell $= (2.88 \overset{˚}{A}) 3 = 23.9 \times 1 0^{- 24} c m^{3}$ .
The volume of 100 g of the metal
$= \frac{m}{ρ} = \frac{100}{7.20} = 13.9 c m^{3}$
Number of unit cells in this volume
$= \frac{13.9 c m ^{3}}{23.9 \times 1 0 ^{- 24} c m ^{3}} = 5.82 \times 1 0^{23}$

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

6.85×102

5.82×1023

4.37×105

2.12×106