# NEET Chemistry The Solid State Questions Solved

BOARD

Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm-3. Use this information to calculate Avogadro's number.

(At. mass of Fe = 55.845 u)

Formula : d = $\frac{Z×M}{{a}^{3}×{N}_{A}}$

For bcc, lattice Z = 2     [bcc = body centred cubic]

7.874 g cm-3

NA

$\therefore$ Avogadro's number, NA = 6.02$×$1023 mol-1

Difficulty Level:

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