How many units cells are there in $1.00 \mathrm{g}$ cube shaped ideal crystal of $AB(M_W=60)$ which has a $\mathrm{NaCl}$ type lattice?

The $AB$ crystal's molar mass is $=60 g/mol$

So, $60 g$of $AB$ will contain $=6\times {10}^{23} atoms$

Now $1 g$ of $AB$$=\frac{6\times {10}^{23}}{60}=0.1\times {10}^{23}\text{ atoms }$ will contain 

One-unit cell of the crystal lattice is made up of 4 atoms because $AB$ again forms a $\mathrm{NaCl}$-type lattice.

So, 

$0.1\times {10}^{23}$ atoms will form  $=\frac{0.1\times {10}^{23}}{4}=0.025\times {10}^{23}$ 

unit cells $=2.50\times {10}^{21}$