JEE CHEMISTRY MCQ: SOLID STATE

1 Hard

1 of 1

Application +4 | -1

An element (atomic mass

100 {~g} / {mol}

) having

b c c

structure has unit cell edge

400 {pm}

. The density of element is (no. of atoms in

b c c, Z=2

)

View answer and explanation

Correct option: C

5.188 {~g} / {cm}^3

Option C (

5.188 {~g} / {cm}^3

) is correct because the density is calculated using the formula

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

, where Z=2 for bcc, M is the molar mass,

N_A

is Avogadro's number, and a is the edge length.

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