# NEET Chemistry The Solid State Questions Solved

BOARD

Iron has a body centred cubic unit cell with a cell edge of 286.65 pm. The density of iron is

7.87 g cm-3. Use this information to calculate Avogadro's number (At. mass of Fe = 56 g mol-1).

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

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

Difficulty Level:

Crack NEET with Online Course - Free Trial (Offer Valid Till September 24, 2019)