Iron oxide FeO, crystallises in a cubic lattice with a unit cell edge length of \(5.0 \AA\). If density of the FeO in the crystal is \(4.0 \mathrm{~g} \mathrm{~cm}^{-3}\), then the number of FeO units present per unit cell is \_\_\_\_ (Nearest integer) Given : Molar mass of Fe and O is 56 and 16 g \(\mathrm{mol}^{-1}\) respectively. \(\mathrm{N}_{\mathrm{A}}=6.0 \times 10^{23} \mathrm{~mol}^{-1}\)

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Solid State

Iron oxide FeO, crystallises in a cubic lattice with a unit cell edge length of

5.0 \overset{˚}{A}

. If density of the FeO in the crystal is

4.0 g cm^{- 3}

, then the number of FeO units present per unit cell is ____ (Nearest integer)

Given : Molar mass of Fe and O is 56 and 16 g

mol^{- 1}

respectively.

N_{A} = 6.0 \times 1 0^{23} mol^{- 1}

Answer: 4

The density of a crystal ties the number of formula units in the cell to its size through

d = \frac{Z M}{N _{A} a ^{3}} .

With

M = 72 g mo l^{- 1}

for

FeO

and edge

a = 5.0 \times 1 0^{- 8} cm

Z = \frac{d N _{A} a ^{3}}{M} = \frac{4.0 \times ( 6.0 \times 1 0 ^{23} ) \times ( 5.0 \times 1 0 ^{- 8} ) ^{3}}{72} .

The cube of the edge is

1.25 \times 1 0^{- 22} c m^{3}

, which makes

Z = 4.17

, so the number of

FeO

units in the unit cell is

4

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Integer answer