Ncert Exercises & Solutions

1.13 Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm⁻³, calculate atomic radius of niobium using its atomic mass 93 u.

It is given that the density of niobium, d = 8.55 g cm⁻³
Atomic mass, M = 93 g mol⁻¹
As the lattice is bcc type, the number of atoms per unit cell, z = 2

We also know that, NA = 6.022 × 10²³ mol⁻¹
Applying the relation:

d = \frac{zM}{a^{3} N_{A}}

\Rightarrow a^{3} = \frac{zM}{{dN}_{A}}

= \frac{2 \times 93 g {mol}^{- 1}}{8.55 g {cm}^{- 3} \times 6.022 \times 10^{23} {mol}^{- 1}}

= 3.612 \times 10^{- 23} {cm}^{3}

So, a = 3.306 × 10⁻⁸ cm
For body-centred cubic unit cell:

r = \frac{\sqrt{3}}{4} a

= \frac{\sqrt{3}}{4} \times 3.306 \times 10^{- 8} cm

= 1.432 \times 10^{- 8} cm

= 14.32 \times 10^{- 9} cm

= 14.32 nm

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