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  4. The number of unit cells in 58.5 g of NaCl is nearly

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Q. The number of unit cells in $58.5 \,g$ of $NaCl$ is nearly

The Solid State

Solution:

$58.5 \,g \,NaCl = 1 $mole $= 6.02 \times 10^{23} Na^+Cl^-$ units
One unit cell contains $4 Na^+Cl^-$ units. Hence number of unit cell present
$ = \frac{6.02\times 10^{23}}{4} = 1.5 \times 10^{23}$