An element with density 2.8 g cm^ –3 forms a f.c.c. unit cell wit… - Vidyadip

Question

An element with density $2.8\ g\ cm^{–3}$ forms a f.c.c. unit cell with edge length $4 \times 10^{–8}\ cm.$ Calculate the molar mass of the element.
$($Given: $N_A = 6.022 \times 10^{23}\ mol ^{–1})$

✓

Answer

Given; $d = 2.8g/cm^3;\ Z = 4;\ a = 4 x 10^{–8}\ cm\ \ N_A= 6.022 x 10^{23}$ per mol
$\text{d}=\frac{\text{Z}\times\text{M}}{\text{a}^{3}\times\text{N}_{A}}$ OR
$\text{M}=\frac{\text{d}\times\text{a}^{3}\times\text{N}_{A}}{\text{z}}$
$\Rightarrow$$\text{M}=\frac{\text{2.8 g cm}^{-3}\times\text{(4}\times\text{10}^{-8}\text{cm})^{3}\times\text{6.022}\times\text{10}^{23}}{\text{4}}$
$M = 2.8 \times 16 \times 10^{–1} \times 6.022 = 26.97$ g/mol.

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