# 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.

(Atomic mass of Fe = 55.84 g mol-1)

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Concept Questions :-

Density/Formula/packing fraction/ Semiconductors

Given :

a = 286.65 pm = 286.65$×$10-10,

d = 7.87 g cm-3,   M = 56 g mol-1

Z = 2         NA= ?

Using formula : d = $\frac{ZM}{{a}^{3}{N}_{A}}$ or NA$\frac{ZM}{{a}^{3}d}$

or NA$\frac{2×56}{\left(286.65×{10}^{-10}{\right)}^{3}×7.87}$

or  NA$\frac{112}{\left(2.87×{10}^{-8}{\right)}^{3}×7.87}$

or  NA$\frac{112}{23.63×7.87×{10}^{-24}}$

or  NA$\frac{112}{185.97×{10}^{-24}}$

or  NA$\frac{112}{18.597}$ $×$ 1023 = 6.022$×$1023

$\therefore$ Avogadro's number NA = 6.022$×$1023

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