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Questions
Gold crystallises into face-centred cubic cells. The edge length of a unit cell is 4.08 × 10–8 cm. Calculate the density of gold. [Molar mass of gold = 197 g mol–1]
When gold crystallizes, it forms face-centred cubic cell. The unit cell edge length is 408 pm. Calculate the density of gold. [Molar mass of gold = 197 g mole–1]
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Solution
Given:
The edge length (a) of the unit cell = 408 pm = 4.08 × 10–8 cm
M = 197 g mol–1
It crystallises in Face-centred cubic cells.
Density (ρ) = `(nM)/(a^3N_A)`
Where n = Number of particles
For FCC, n = 4
M = Molar Mass
a = Edge length
NA = 6.022 × 1023
ρ = `(197 xx 4)/((4.08 xx 10^-8)^3 xx 6.022 xx 10^23)`
ρ = `788/(67.92 xx 10^-24 xx 6.022 × 10^23)`
ρ = `788/(408.99 xx 10^-1)`
ρ = 1.927 × 10−1
ρ = 19.27 g cm–3
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