Question

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

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

Answer 1

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

Overall $4$ units $NaCl$ are present in one cell

$n$ of moles = $\frac{1}{(23+35.5)}$

$=\frac{1}{58.5}$

$n$ of molecular = $6.022\times {10}^{23}\times \frac{1}{58.5}$

$\therefore n$ of unit cells =$\frac{molecules}{4}$

$=\frac{6.022\times {10}^{23}}{4\times 58.5}$

$=\frac{602.2\times {10}^{21}}{234}$

$=2.57\times {10}^{21}$