Question

Crystallize $CsCl$ has density 3.988 g $c{m}^{-3}.$ The volume occupied by a single $CsCl$ ion pair in the crystal will be:

• A. $7.014\times {10}^{-3}c{m}^3$
• B. $7.014\times {10}^{-23}c{m}^3$
• C. $1.014\times {10}^{-3}c{m}^3$
• D. $1.542\times {10}^{-5}c{m}^3$

Answer 1

Answer: (B)

$7.014\times {10}^{-23}c{m}^3$

Theoretical density of crystal, $\rho =\frac{nM}{N_o{a}^3}g/c{m}^3$

Now, $CsCl$ crystallizes in such a way that $C{l}^{-}$ atoms are at corners and $C{s}^{+}$ ion is at the body center of the cubic unit cell. So, number of formula units of $CsCl$ is $1$. So, there will be total $2$ ions in a unit cell.

Given, $\rho =3.988g/c{m}^3;M=168.5g;n=1$

So, $a^3=\frac{nM}{N_0.\rho }$

$a^3=\frac{1\times 168.5}{6.022\times {10}^{23}\times 3.988}$

$a^3=7.014\times {10}^{23}c{m}^3$

Volume of the unit cell $=$ volume occupied by single $CsCl$ pair.

$=7.01\times {10}^{-23}c{m}^3$

Hence, Option "B" is the correct answer.