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Chemistry Question on Unit Cells

The number of atoms in 100 g100\,g of an fccfcc crystal with density, d=10 g/cm3d = {10\, g/cm^3} and cell edge equal to 100 pm100\,pm, is equal to

A

2×10252 \times 10^{25}

B

1×10251 \times 10^{25}

C

4×10254 \times 10^{25}

D

3×10253 \times 10^{25}

Answer

4×10254 \times 10^{25}

Explanation

Solution

Mass(m)=100 g;Density(d)=10 g/cm3{Mass} (m) = {100\, g; Density} (d) = 10\, {g/cm^3}
and length (l)=100 pm=100×10−12 m {length } \,(l) ={ 100\, pm = 100 \times 10^{-12}\, m }
=100×10−10cm= 100 \times 10^{-10}{ cm}
We know that volume of the unit cell
=(l)3=(100×10−10 cm)3=10−24cm3= (l)^3 ={ (100 \times 10^{-10} \, cm)^3 = 10^{-24}cm^3} and volume of 100 g of element
=MassDensity=10010=10 cm3= {\frac{Mass}{Density} = \frac{100}{10} = 10 \, cm^3}
Therefore, number of unit cells = 1010−24=1025\frac{10}{10^{-24}} = 10^{25}
Since each fcc cube contains 4 atoms,
therefore total number of atoms in 100 g100\, g
=4×1025= 4 \times 10^{25}