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Chemistry Question on The solid state

An element crystallises in a structure having a fccfcc unit cell of an edge 200 pm200\,pm. If 200 g200\,g of this element contains 24×102324 \times 10^{23} atoms then its density is

A

41.66 g cm−341.66\,g\,cm^{-3}

B

313.9 g cm−3313.9\,g\,cm^{-3}

C

8.117 g cm−38.117\,g\,cm^{-3}

D

400 g cm−3400\,g\,cm^{-3}

Answer

41.66 g cm−341.66\,g\,cm^{-3}

Explanation

Solution

Molar mass of the element =20024×1023×6.023×1023=\frac{200}{24\times10^{23}}\times6.023\times10^{23} =50.19 g mol−1=50.19\,g\,mol^{-1} For fcc,Z=4fcc, Z=4, V=a3=(200×10−10)3V=a^{3}=\left(200\times10^{-10}\right)^{3} d=Z×MNA ×Vd=\frac{Z\times M}{N_{A}\,\times V} =4×50.196.023×1023×(200×10−10)3=\frac{4\times50.19}{6.023\times10^{23}\times\left(200\times10^{-10}\right)^{3}} =41.66 g cm−3=41.66\,g\,cm^{-3}