Question

In a solid substance edge length of unit cell is 405 pm. density of solid is 10.9 gram/cm^3 and molar mass of substance is 109 gram then find radius of atom which form this unit cell in pm ?
​

Answer 1

Answer:

Given:-

• Length of unit cell (l) = 405 pm or 405×10^-10cm

• Density (ρ) = 10.9 g/cm³

• Molar mass (M) = 109 g

Formula:

Density of unit cell,

$\bf \large \rho = \frac{Z × M }{N_o× {a}^{3} }$

Substitute the value in Equation,

$\bf \large \longmapsto \: 10.9 = \frac{Z \times 109}{6.02 \times {10}^{23} \times (405 \times {10}^{10}) ^{3} }$

$\bf \large \longmapsto \: 10.9 = \frac{Z \times 109}{6.02 \times {10}^{23} \times 66430125 \times {10}^{ - 30} } \\ \\ \bf \large \longmapsto \: 10.9 = \frac{Z \times 2.72 \times {10}^{ - 7} }{ {10}^{ - 7} } \\ \\ \bf \large \longmapsto \: Z = \frac{10.9}{2.72} \\ \\ \bf \large \longmapsto \: Z =4$

We know that,

$\sf \large { r = \frac{ \sqrt{2}a }{4} }$

Substitute value of a in the equation,

$\bf \large : \longmapsto \: r = \frac{ \sqrt{2} \times 405}{4} \\ \\ \bf \large : \longmapsto \: r =143.2 \: \: pm$

$\bf \large \underline \blue{Radius \: \: of \: \: atom \: \: (r) \: \: = \: \: 143.2 \: \: pm}$