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प्रश्न
An element with molar mass 27 g/mol forms a cubic unit cell with edge length of 405 p.m. If the density of the element is 2.7 g/cm3, what is the nature of the cubic unit cell?
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उत्तर
Given:
Edge length (a) = 405 pm = 4.05 × 10−8 cm
Molar mass = 27 g mol−1
Density (ρ) = 2.7 g/cm3 = 2.7 g cm−3
To find: Nature of cubic unit cell
Formula: Density (ρ) = `(M n)/(a^3 N_A)`
Calculation: From the formula,
Density, ρ = `(M n)/(a^3 N_A)`
∴ `2.7 "g cm"^-3 = (27 "g mol"^-1 xx "n")/((4.05 xx 10^-8)^3 "cm"^3 xx 6.022 xx 10^23 "atom mol"^-1)`
∴ n = `(2.7 "g cm"^-3 xx (4.05 xx 10^-8)^3 "cm"^3 xx 6.022 xx 10^23 "atom mol"^-1)/(27 "g mol"^-1)`
= 4.00
∴ Number of atoms in unit cell = 4
Since the unit cell contains 4 atoms, it has a face-centred cubic (fcc) or ccp structure.
The nature of the given cubic unit cell is face-centred cubic (fcc) or ccp unit cell.
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