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Chemistry Question on The solid state

A solid having density of 9×103  kg  m−39 \times 10^3 \; kg \; m^{-3} forms face centred cubic crystals of edge length 2002200 \sqrt{2} pm. What is the molar mass of the solid ? [Avogadro constant ≃6×1023 mol−1,π≃3]\simeq 6 \times 10^{23} \, mol^{-1}, \pi \simeq 3]

A

0.0216  kg  mol−1{0.0216 \; kg \; mol^{-1}}

B

0.0305  kg  mol−1{0.0305 \; kg \; mol^{-1}}

C

0.4320  kg  mol−1{0.4320\; kg \; mol^{-1}}

D

0.0432  kg  mol−1{0.0432\; kg \; mol^{-1}}

Answer

0.0305  kg  mol−1{0.0305 \; kg \; mol^{-1}}

Explanation

Solution

Formula for density,
Density, d=Z×Ma3×NAd =\frac{ Z \times M }{ a ^{3} \times N _{ A }}
where,
d=d = density of the unit cell
M=M = Molar mass of the molecule
a3=a^{3}= volume of the unit cell
NA=N _{ A }= Avogadro number
Here, for FCC unit cell, Z=4Z = 4
Subsituting the values we get,
9×103=4×M(2002×10−12)3×6.0×10239 \times 10^{3}=\frac{4 \times M }{\left(200 \sqrt{2} \times 10^{-12}\right)^{3} \times 6.0 \times 10^{23}}
⇒M=0.0305 kg.mol−1\Rightarrow M =0.0305 \,kg .mol ^{-1}