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# Aluminum crystallizes in a cubic close-packed structure. Its metallic radius is 125 pm. (i) What is the length of the side of the unit cell? (ii) How many unit cells are there in $1.00{\mathrm{cm}}^{3}$ of aluminum?

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Solution

## Step 1: Given dataThe radius of the metal is $125\mathrm{pm}$Volume of Aluminium given is $1.00{\mathrm{cm}}^{3}$Step 2: Calculating the lengthWe know that for closed-packed structures:$\mathrm{a}=2\sqrt{2}\mathrm{r}\phantom{\rule{0ex}{0ex}}⇒\mathrm{a}=2\sqrt{2}×125\phantom{\rule{0ex}{0ex}}⇒\mathrm{a}=354\mathrm{pm}$Step 3: Calculating the number of unit cellAs we have the radius (r) as $125\mathrm{pm}$ and the length (a) as $354\mathrm{pm}$$\mathrm{Volume}\mathrm{of}\mathrm{unit}\mathrm{cell}={\mathrm{a}}^{3}\phantom{\rule{0ex}{0ex}}\mathrm{V}={\left(354\mathrm{pm}\right)}^{3}\phantom{\rule{0ex}{0ex}}\mathrm{V}=4.4×{10}^{-23}{\mathrm{cm}}^{3}\left[\because 1\mathrm{pm}={10}^{-10}\mathrm{cm}\right]$$\therefore$Number of unit cells in $1.00{\mathrm{cm}}^{3}$ of aluminium is:$\frac{1.00}{4.4×{10}^{-23}}=2.27×{10}^{22}$Therefore, length of the side of the unit cell is 354 pm and Number of unit cells in $1.00{\mathrm{cm}}^{3}$ of aluminium is $2.27×{10}^{22}$  Suggest Corrections  0      Explore more