Question

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

• A. $3.6 \times 10^{18}$
• B. $3.6 \times 10^{15}$
• C. $2.34 \times 10^{23}$
• D. $6.02 \times 10^{24}$

Answer 1

Answer: (A)

In a unit cell of KBr, there are four formula units of $KBr$ therefore,
Density
$( \rho )=\frac{4 \times 119}{6.023 \times 10^{23} \times a ^{3}}=2.75$
$a^{3} =2.873 \times 10^{-22} cm ^{3}$
$=2.778 \times 10^{-19} mm ^{3}$
Number of unit cells per $mm ^{3}$
$=\frac{10^{19}}{2.873}=3.5 \times 10^{18}$