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The cubic unit cell of aluminium has an edge length of 405 pm. If its density is 2.7 gcm$^{-3}$, what is the type of unit cell? (Note: Molar mass = 27 g/mol)

$\begin{array}{1 1} Face\;centered \\ Primitive \\ Body\;centered \\ Diamond\;shaped \end{array}$

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Answer: Face centered cube
The density is given by the formula $\large\frac{z\;M}{a^3\;N_A}$
Given $a = 405 \;ppm, \;$ Density $\rho = 27\;g\;cm^{-3}$ and Molar mass $= 27 \;g/mol$, we need to determine $z$.
Avagadro's number $N_A = 6.022 \times 10^{23} /\; mol$
$\Rightarrow z = \large\frac{\rho a^3 N_A}{M} $$=\large\frac{27 \times (405\times10^{-10})^3 \times 6.022 \times 10^{23}}{27}$$ = 4 $
$\Rightarrow z = 4 \rightarrow$ that the unit cell is a face-centered cube.
answered Jul 15, 2014 by balaji.thirumalai
edited Jul 15, 2014 by balaji.thirumalai
 

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