Question

How many unit cells are there in $1.00$ gm cube shaped ideal crystal $AB(MF=60)$ having $NaCl$ type lattice?

• A. $6.02\times {10}^{23}$
• B. $2.50\times {10}^{21}$
• C. $1.00\times {10}^{22}$
• D. $6.02\times {10}^{24}$

Answer 1

Answer: (B)

$2.50\times {10}^{21}$

In $NaCl$ type lattice,

$N{a}^{+}⟶$present octahedral voids

Thus, no. of $N{a}^{+}$ ions=4

$C{l}^{-}$ is present in fcc structure.

Thus, no. of $C{l}^{-}$ ions=4

1 unit cell contains four $NaCl$ molecules.

Thus, 1 unit cell contains 4$N{a}^{+}$ ion and 4$C{l}^{-}$ ions.

No. of moles in 1 g of crystal of AB=$\frac{1}{60}$moles

1 mole contains=$6.023\times {10}^{23}$ ions

$\frac{1}{60}$ moles contains=$\frac{6.023\times {10}^{23}}{60}$ ions or molecules.

1 unit cell has=8 ions or 4 molecules

or 1 ion=$\frac{1}{8}$ unit cell

$\frac{6.023\times {10}^{23}}{60}$ ions=$\frac{1}{8}\times \frac{6.023\times {10}^{23}}{60}$ unit cell

$\frac{6.023\times {10}^{23}}{60}$ molecules=$\frac{1}{4}\times \frac{6.023\times {10}^{23}}{60}$ unit cell

Thus, 1 g of crystal contains $2.5\times {10}^{21}$ unit cell.