Question

An element crystallizes in a structure having FCC unit cells of an edge length 200 pm. Calculate the density (in ${\rm{g}}\,{\rm{c}}{{\rm{m}}^{ - 3}}$) if 200 grams of it contains $24 \times {10^{23}}$ atoms.

A.41.6
B.42.6
C.43.6
D.44.6

Answer 1

A face centred cubic (FCC) unit cell has atoms at centre of all faces and at all corners of the cube. To calculate density of the element, we have to use the formula of density, that is, $d = \dfrac{{zM}}{{{a^3}{N_A}}}$, where, d is density, z is the number of atoms, M is molar mass, a is edge length and ${N_A}$ is Avogadro’s number.

Complete step by step answer:
To use the formula of density, we need the values of z , M , a and ${N_A}$. For FCC unit cell, number of atoms (z) is 4, the value of edge length (a) is given as 200 pm, value of ${N_A}$ is $6.022 \times {10^{23}}$.

Now, we have to calculate the molar mass of the element.

$24 \times {10^{23}}$ atoms present in=200 g of the element

By using unitary method, $6.022 \times {10^{23}}$ atoms present in$= \dfrac{{200 \times 6.022 \times {{10}^{23}}}}{{24 \times {{10}^{23}}}} = 50.18\,{\rm{g}}$of element

So, the molar mass (M) of the element in 50.18 ${\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$.

Given that, we have to calculate density in ${\rm{g}}\,{\rm{c}}{{\rm{m}}^{ - 3}}$, that means, we have to convert edge length to cm from pm. The conversion factor is $1\,{\rm{pm}} = {10^{ - 10}}\,{\rm{cm}}$.

$\Rightarrow 200\,pm = 200 \times {10^{ - 10\,}}{\rm{cm}}$

Now, we have to put all the values ( z=4, M= 50.18 ${\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$
a=$200 \times {10^{ - 30}}\,{\rm{c}}{{\rm{m}}^{\rm{3}}}$ and ${N_A}$ = $6.022 \times {10^{23}}$) in the formula of density.

$d = \dfrac{{zM}}{{{a^3}{N_A}}}$

$\Rightarrow d = \dfrac{{4 \times 50.18\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}\,}}{{{{\left( {200 \times {{10}^{ - 10}}\,{\rm{cm}}} \right)}^3}\, \times 6.022 \times {{10}^{23}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}}}$

$\Rightarrow d = \dfrac{{200.72\,{\rm{g}}}}{{8 \times {{10}^6} \times {{10}^{ - 30}} \times 6.022 \times {{10}^{23}}\,{\rm{c}}{{\rm{m}}^{\rm{3}}}}}$

$\Rightarrow d = \dfrac{{200.72\,\,{\rm{g}}}}{{48.176 \times {{10}^{ - 1}}\,{\rm{c}}{{\rm{m}}^{\rm{3}}}}}$

$\Rightarrow d = 41.67\,{\rm{g}}\,{\rm{c}}{{\rm{m}}^{ - 3}}$

Therefore, the density of the element is $41.67\,{\rm{g}}\,{\rm{c}}{{\rm{m}}^{ - 3}}$. Hence, correct answer is A.

Note:
The smallest repeating unit in a crystal is termed as unit cell. There are three types of unit cell, such as body centred cubic (BCC) unit cell, primitive cubic unit cell and face centred cubic (FCC)unit cell. The number of atoms present in FCC unit cells is 4, the number of atoms present in BCC unit cells is 2 and the number of atoms in primitive unit cells is 1.