Question

How many unit cells are present in a cube-shaped ideal crystal of $NaCl$ of mass $1\text{g}$?

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

Answer 1

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

Molar mass of $NaCl$ = 58.5 g
$1\text{g}NaCl\text{contains}=\frac{1}{58.5}\text{moles of}NaCl$
or,
$1\text{g}NaCl\text{contains}=\frac{1}{58.5}\times 6.02\times {10}^{23}\text{molecules of}NaCl$

One unit cell has '4' $NaCl$ molecules i.e., 1 molecule has $\frac{1}{4}$ unit cell
$\therefore$
$\text{Number of unit cells in}\frac{1}{58.5}\times 6.02\times {10}^{23}\text{molecules of}NaCl$
$=\frac{6.02\times {10}^{23}}{4\times 58.5}=2.57\times {10}^{21}$