Question

$Csl$ lattice has structure of two interpenetrating simple cubic structure, the icons being at the centre of lattice of $C{s}^{+}$ ions. The unit cell side length $=4.562\mathring{A}$. Calculate density $(Ca=132.9,I=126)$:

• A. $4.002gc{m}^{-3}$
• B. $4.996gc{m}^{-3}$
• C. $4.546gc{m}^{-3}$
• D. $4.732gc{m}^{-3}$

Answer 1

The correct option is C $4.546gc{m}^{-3}$

$\to Z=1$ (for $CsCl$ type), i.e., one formula unit of $CsCl$ present in the $C{s}^{+}$ at body centre, $C{l}^{-}$ at body corners.

$M_A=258.9$

$a=4.562\mathring{A}$

$\rho =\frac{Z\times M_A}{N_A\times a^3}$

$⟹\rho =\frac{1\times 258.9}{6.023\times {10}^{23}\times 94.94\times {10}^{-24}}$ $g/c{m}^3$

${\rho }_{CsCl}=4.546$ $g/c{m}^3$