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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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