Question

NA= 6.023×10^23
M = 58.5×10^-3
Z= 4
D= 2.17×10^3
a=?

(plz solve this numerical)​

Answer 1

Given:

Na = 6.023 × 10²³ mol⁻¹

M = 58.5 × 10⁻³ gmol⁻¹

Z = 4

D = 2.17 × 10³ kgm⁻³

To find:

Edge length

Solution:

To determine edge length formula used is -

   $D = \frac{Z \times M }{a^3 \times N_a}$

or,   $a^3 = \frac{Z \times M}{D \times N_a}$

here, a = edge length

        Z = no. of atoms per unit cell

        M = atomic mass

         D = density

        Na = Avogadro's number

now, substitute all the values we get,  

      $a^3 = \frac{4 \times 58.5 \times 10^-^3}{2.17 \times 10^3 \times 6.023 \times 10^2^3}$

      $a^3 = \frac{243 \times 10^-^3 }{13.069 \times 10^2^6}$

     a³ = 18.59 × 10⁻²⁹

     a = $\sqrt[3]{18.59 \times 10^-^2^9}$

    a = 1.23 × 10⁻¹⁴ m

Therefore, the edge length is 1.23 × 10⁻¹⁴m.