Question

How many unit cells are present in a cube shaped ideal crystal of $NaCl$ of mass $1.00$ g?

• A. $2.57\times {10}^{21}$
• B. $5.14\times {10}^{21}$
• C. $1.28\times {10}^{21}$
• D. $1.71\times {10}^{21}$

Answer 1

The correct option is A $2.57\times {10}^{21}$

Mass of one unit cell= $V\times d$ $=a^3\times d$, where $V$ is volume and $d$ is density.

Also, $\text{density}=\frac{n\times \text{formula mass}}{a^3\times \text{Avogadro's number}}$

where n is the no. of NaCl formula unit per unit cell
$\therefore$ Mass of one unit cell $=\frac{a^3\times n\times \text{formula mass}}{a^3\times \text{Avogadro's number}}$ $=\frac{n\times \text{formula mass}}{N_a}$

$C{l}^{-}$ ions are present at the FCC lattice, i.e. corners and face centers. $N{a}^{+}$ ions are present at all the octahedral voids.

No. of $N{a}^{+}$ ions per unit cell = 4

No. of $C{l}^{-}$ per unit cell - 4

No. of formula unit of NaCl per unit cell= 4

Formula mass for NaCl= $M_{Na}+M_{Cl}=23+35.5=58.5$
$=\frac{4\times 58.5}{6.02\times {10}^{23}}$

$=38.87\times {10}^{-23}$ g
$\therefore$ Number of unit cells in $1$ g $=\frac{1}{38.87\times {10}^{-23}}$ $=2.57\times {10}^{21}$

Hence, the correct option is A.