CsCl has a cubic cell with Cs+ at the body centre and Cl- at the corners, the ions touching along the body diagonal. Taking r(Cs+)=169 pm, r(Cl-)=181 pm, M=168.5 g mol^-1, sqrt3=1.732 and N_A=6.022x10^23 mol^-1, calculate the density in g cm^-3 (two decimals).

JEE Advanced levelNumerical answerDifficulty 3 of 3ss036

CsCl has a cubic cell with $Cs X^{+}$ at the body centre and $Cl X^{-}$ at the corners, the ions touching along the body diagonal. Taking $r (Cs X^{+}) = 169 pm$ , $r (Cl X^{-}) = 181 pm$ , $M = 168.5 g mo l^{- 1}$ , $3 = 1.732$ and $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , calculate the density in $g c m^{- 3}$ (two decimals).

Numerical answer type, answer in g/cm^3, tolerance 0.02.

Ions touch along the body diagonal, so

3 a = 2 (r_{+} + r_{-}) = 700 pm

and

a = 700/1.732 = 404.2 pm

. The cell has

Z = 1

formula unit.

ρ = \frac{168.5}{6.022 \times 1 0 ^{23} \times ( 4.0416 e - 08 ) ^{3}} = 4.24 g c m^{- 3}