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Question
An element with molar mass 2.7 × 10-2 kg mol-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 103 kg m−3, what is the nature of the cubic unit cell?
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Solution
It is given that density of the element, d = 2.7 × 103 kg m−3
Molar mass, M = 2.7 × 10−2 kg mol−1
Edge length, a = 405 pm = 405 × 10−12 m
= 4.05 × 10−10 m
It is known that, Avogadro’s number, NA = 6.022 × 1023 mol−1
Applying the relation,
`d = (z,m)/(a^3.N_A)`
`z= (d.a^3N_A)/M`
`=(2.7xx10^3kgm^(-3)xx(4.05xx10^(-10 m))^3 xx6.022xx10^23 mol^(-1))/(2.7xx10^(-2)kg mol^(-1))`
= 4.004
= 4
This implies that four atoms of the element are present per unit cell. Hence, the unit cell is face-centred cubic (fcc) or cubic close-packed (ccp).
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