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Question: he density of potassium bromide crystal is \[\,2.75g/c{m^3}\,\] and the length of an edge of the uni...

he density of potassium bromide crystal is 2.75g/cm3\,2.75g/c{m^3}\, and the length of an edge of the unit cell is 654pm654pm .The unit cell of KBrKBr is one of the three types of cubic unit cell. How many formula units of KBrKBr are there in a unit cell? [KBr=119][KBr = 119]
A.22
B.44
C.66
D.88

Explanation

Solution

The answer for this question can be obtained by using the formula to calculate density of the unit cell. A unit cell's density is given as the ratio of the unit cell's mass and volume. The mass of a unit cell is proportional to the product of a unit cell's number of atoms and a unit cell's mass of each atom.
Formula used:
ρ=Z×MNA×a3\rho = \dfrac{{Z \times M}}{{{N_A} \times {a^3}}}
Where ZZ= Number of formula units
MM= Molar mass of KBrKBr
NA{N_A}= Avogadro number
aa = Edge length

Complete step by step answer:
Let us analyze the given data;
Density ρ=2.75g/cm3\rho = 2.75g/c{m^3}
Edge length a=654pma = 654pm which is equal to 654×1010cm654 \times {10^{ - 10}}cm
Molar mass of KBrKBr =119g/mol = 119g/mol
We Know that formula for calculating the density of unit cell is,
ρ=Z×MNA×a3\rho = \dfrac{{Z \times M}}{{{N_A} \times {a^3}}}
From this we can arrange the formula for calculating the number of formula units ZZ, that is,
Z=ρ×NA×a3MZ = \dfrac{{\rho \times {N_A} \times {a^3}}}{M}
=2.75×6.022×1023×(654×1010)3119= \dfrac{{2.75 \times 6.022 \times {{10}^{23}} \times {{(654 \times {{10}^{ - 10}})}^3}}}{{119}}
=3.9= 3.9 which is approximately equal to 44.
So, the right answer is option B that is 44.

Additional information:
Formula unit is the simplest ratio at which an ionic compound exists. Or simply we can say its empirical formula. The number of formula units is the number of atoms per unit cell which is denoted by ZZ.

Note:
Note that while doing the calculations 654pm654pm should be changed to cmcm since the value of density in question is given in g/cm3g/c{m^3}, Otherwise the answer would be incorrect. 1centimetre\,1\,centimetre\, is equal to 10000000000picometre\,10000000000\,picometre\, . In addition, note that not all the particles in the shape of the unit cell are complete. Its corner particles will still add up to one whole particle for the fractional particles in the unit cell, its face particles (for face-centered lattices) add up to three whole particles, and one for the base particles.