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A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g $cm^{-3}$. The molar mass of the metal is:
A
20 g $mol^{-1}$
B
40 g $mol^{-1}$
C
30 g $mol^{-1}$
D
28 g $mol^{-1}$
Detailed Solution
$\rho = \frac{ZM}{N_A\times a^3}$; for fcc, Z = 4; a = 404 pm = $404\times10^{-10}$ cm
$2.72 = \frac{4\times M}{6.02\times10^{23}\times(404\times10^{-10})^3}$
M ≈ 27 g $mol^{-1}$, closest to 28 g $mol^{-1}$.
$2.72 = \frac{4\times M}{6.02\times10^{23}\times(404\times10^{-10})^3}$
M ≈ 27 g $mol^{-1}$, closest to 28 g $mol^{-1}$.