Question

The density of KBr is 2.75 g/cm*-3 .the length of edge of the unit cell is6.54 pm .prodict the type of cubic lettice to which the unit cell KBr belong .[atomic mass of K= 39 and Br = 80].
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Answer 1

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Answer 2

Answer : The type of cubic lattice is, Face centered cubic unit cell(FCC), Z = 4

Solution : Given,

Edge length = 654 pm = $654\times 10^{-10}cm$     $(1pm=10^{-10}cm)$

Density of KBr = $2.75g/cm^3$

Molar mass of KBr = 39 + 80 = 119 g/mole

Density of the crystal lattice is given by the formula :

$\rho=\frac{Z\times M}{N_A\times a^3}$

where,

Z = number of atoms in crystal lattice

M = Atomic mass

a = Edge length of the crystal

$\rho=Density\\N_A=\text{Avogadro's Number}=6.022\times 10^{23}$

Now put all the given values in the above formula, we get

$2.75g/cm^3=\frac{Z\times (119g/mole}{(6.022\times 10^{23})\times (654\times 10^{-10}cm)^3}$

$Z=3.89=4atoms$

Therefore, the type of unit cell is, FCC.