Question

An element with molar mass $$27 g mol^{-1}$$ forms a cubic unit cell with edge length $$4.05\times 10^{-8}cm$$. If its density is $$2.7 g cm^{-3}$$, what is the nature of the cubic unit cell?

Solution

Given:Molar mass of the given element, $$M=27 g\;mol^{-1}=0.027 kg\;mol^{-1}$$Edge length, $$a=4.05\times 10^{-8}cm=4.05\times 10^{-10}m$$Density, $$d=2.7g\;cm^{-3}=2.7\times 10^{-3}kg\;m^{3}$$We know that $$\displaystyle d=\frac{Z\times M}{a^{3}\times N_{A}}$$Where, Z is the number of atoms in the unit cell and $$N_{A}$$ is the Avogadro number. Thus,$$\displaystyle Z=\frac{d\times a^{3}\times N_{A}}{M}$$$$\displaystyle Z=\frac{2.7\times 10^{3}(4.05\times 10^{-10})^{3}\times 6.022\times ^{23}}{0.027}$$                                                           $$(N_{A}=6.022\times 10^{23})$$$$Z=4\;or\;z\approx 4(fcc)$$Since, the number of atoms in the unit cell is 4, the given cubic unit cell has face-centred cubic (fcc) or cubic-close packed (ccp) structure.Chemistry

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