Question

How many unit cells are present in a cube-shaped ideal crystal of NaCl of mass $1.00$ g? [ Atomic masses : Na = $23$, Cl = $35.5$]

• 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

Answer: (A)

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

Molar mass of $NaCl=58.5$

$1g$ of $NaCl$ contains $=\frac{1}{58.5}$ moles of $NaCl$

$1g$ of $NaCl$ contains $\frac{1}{58.5}\times 6.022\times {10}^{23}$ molecule of $NaCl$ in cube-shaped ideal crystal of $NaCl$ (Rock salt type)
4 molecule of $NaCl$ are present as ${Cl}^{-}$ present occupy the FCC lattice$=4$ ions of ${Ce}^{-}$

${Na}^{+}$ occupy the octanedral site $=4$ ions of ${Na}^{+}$

$1$ unit cell contain $4$ molecules of $NaCl$

$1$ molecule contain $\frac{1}{4}$ unit cell

$\frac{1}{58.5}\times 6.022\times {10}^{23}$ molecule contains$=\frac{1}{58.5}\times \frac{6.022\times {10}^{23}}{4}$ unit cell

Thus

$1g$ of $NaCl$ contain$=2.57\times {10}^{21}$ unit cell.