Question

${m{CsBr}}$ crystallises in a body centred cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of ${m{Cs}} = 133$ and that of ${m{Br}} = 80$ amu and Avogadro number being $6.02 	imes {10^{23}}{m{mo}}{{m{l}}^{ - 1}},$ the density of ${m{CsBr}}$ is :

Accepted Answer

$4.25\;{\rm{g}}/{\rm{c}}{{\rm{m}}^3}$

Answer

$8.25\;{\rm{g}}/{\rm{c}}{{\rm{m}}^3}$

Answer

$42.5\;{\rm{g}}/{\rm{c}}{{\rm{m}}^3}$

Answer

$0.425\;{\rm{g}}/{\rm{c}}{{\rm{m}}^3}$

NEET UG CHEMISTRY MCQ: SOLID STATE

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