Question

$KBr$ crystallises in $NaCl$ type of crystal lattice and its density is $2.75g/c{m}^3$. Number of unit cells of $KBr$ present in a 1.00 $m{m}^3$ grain of $KBr$ are:

• A. $3.6\times {10}^{18}$
• B. $3.6\times {10}^{16}$
• C. $2.34\times {10}^{23}$
• D. $6.02\times {10}^{24}$

Answer 1

The correct option is A $3.6\times {10}^{18}$

In a unit cell of $KBr,$ there are four formula units of $KBr$ therefore,

Density $\left(\rho \right)$ $=\frac{4\times 119}{6.023\times {10}^{23}{a}^3}=2.75$

$a^3=2.873\times {10}^{-22}c{m}^3$

$=2.873\times {10}^{-19}m{m}^3$

Number of units cells per $m{m}^3$

$\frac{1}{2.873\times {10}^{-19}}=\frac{{10}^{19}}{2.873}=3.5\times {10}^{18}$

Option A is correct.