Question

How many unit cells are present in a cube-shaped ideal crystal of $NaCl$ of mass $1.00\text{g}$?
$(N_A=6\times {10}^{23})$

• 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 $2.57\times {10}^{21}$

$\text{mass (m) = density}\left(\rho \right)\times \text{volume}=1.00\text{g}$
$\text{Molecular weight (M) of NaCl}=23+35.5=58.5u$
$\text{Number of unit cells present in a cube shaped crystal of}NaCl\text{of mass}1.00\text{g}$
$=\frac{\rho \times a^3\times N_A}{M\times Z}=\frac{m\times N_A}{M\times Z}=\frac{1\times 6\times {10}^{23}}{58.5\times 4}$
In $NaCl$, each unit cell has $4$ $NaCl$ units. Hence, $Z=4$
$\therefore \text{Number of unit cells}=0.02564\times {10}^{23}=2.56\times {10}^{21}\text{unit cells}$