How many unit cells are present in a cube shaped ideal crystal of $\mathrm{NaCl}$ of mass lg?
How many unit cells are present in a cube shaped ideal crystal of $\mathrm{NaCl}$ of mass lg?
$5.14 \times 10^{21}$
$1.28 \times 10^{24}$
$1.71 \times 10^{21}$
$2.5 \times 10^{21}$
Solution
Molar mass of $\mathrm{NaCl}=58.5$
lg of $\mathrm{NaCl}=\frac{1}{58.5}$ mole of $\mathrm{NaCl}$
$1 \mathrm{~g}$ of $\mathrm{NaCl}=\frac{1}{58.5} \times 6.022 \times 10^{23}$ molecules of $\mathrm{NaCl}$
4 molecules of $\mathrm{NaCl}$ are present as $\mathrm{Cl}^{-}$present occupy fcc lattice $=4$ ions of $\mathrm{Cl}^{-}$
occupy $\mathrm{Na}^{+}$octahedral site $=4$ ions of $\mathrm{Na}^{+}$
1 unit cell contain 4 molecules of $\mathrm{NaCl}$
1 molecule contain $\frac{1}{4}$ unit cell.
$\frac{1}{58.5} \times 6.022 \times 10^{23}$ molecule contains
$
=\frac{1}{58.5} \times \frac{6.022 \times 10^{23}}{4} \text { unit cell }
$
Thus, $1 \mathrm{~g}$ of $\mathrm{NaCl}$ contain $=2.57 \times 10^{21}$ unit cell