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Question
An element has atomic mass 93 g mol−1 and density 11.5 g cm–3. If the edge length of its unit cell is 300 pm, identify the type of unit cell. (NA = 6.023 × 1023 mol−1)
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Solution
Given: M = 93 g mol−1
ρ = 11.5 g cm−3
a = 300 pm = 300 × 10−10 cm = 3 × 10−8 cm
NA = 6.023 × 1023 mol−1
Formula: `rho = (Z xx M)/(N_A xx a^3)`
`Z = (rho xx N_A xx a^3)/M`
`Z = (11.5 xx (3 xx 10^-8)^3 xx6.023 xx 10^23)/93`
`Z = (11.5 xx 27 xx 10^-24 xx 6.023 xx 10^23)/93`
`Z = (11.5 xx 162.621 xx 10^-1)/93`
`Z = (1870.14 xx 10^-1)/93`
`Z = 187.014/930`
Z = 2.01 (approx.)
As the number of atoms present in given unit cells is coming nearly equal to 2, hence the given unit cell is a body-centred cubic unit cell (bcc).
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