In a CsCl-type crystal the ions touch along the body diagonal of the cubic cell. Take r(Cs^+)=169 pm, r(Cl^-)=181 pm, sqrt3=1.732, atomic masses Cs =133, Cl =35.5 and N_A=6.022x10^23 mol^-1. Calculate the density in g cm^-3 (two decimals).

JEE Advanced levelNumerical answerDifficulty 3 of 3ss070

In a CsCl-type crystal the ions touch along the body diagonal of the cubic cell. Take $r (C s^{+}) = 169 pm$ , $r (C l^{-}) = 181 pm$ , $3 = 1.732$ , atomic masses Cs $= 133$ , Cl $= 35.5$ 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.

Body diagonal:

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

, so

a = 2 \times 350/1.732 = 404.2 pm

. With

Z = 1

ρ = \frac{168.5}{6.022 \times 1 0 ^{23} \times ( 4.042 \times 1 0 ^{- 8} ) ^{3}} = 4.238 \approx 4.24 g c m^{- 3}