Question

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An element A (atomic mass 150) having bcc structure has unit cell edge length of 400 pm.
Calculate the density and number of unit cells in 7.78 g of A.​

Answer 1

Given:

Atomic mass = 150g/mol , Edge length(a) = 400pm, Given mass = 7.78g

To find:

Density and number of unit cells

Solution:

• We know for a BCC structure Number of Atoms per unit cell (Z) = 2

• Now density(d) =   $\frac{ZM}{a^{3}N_{A} }$    ⇒ d =   $\frac{2*150}{(400*10^{-10})^{3} *(6.022*10^{23}) }$ = 7.8125g/cm³

         d = 7.8g/cm³

• Now for number of unit cell,

• No. of atoms = (given mass / molar mass)×Na

                      = (7.78÷150)×(6.022×$10^{23}$) ⇒ 3.12×$10^{22}$ atoms

 

• No. of unit cell =  $\frac{No. of atoms }{No. of atoms per unit cell(Z)}$    = $\frac{3.12*10^{22} }{2}$ = 1.56×$10^{22}$

• So density(d) = 7.8g/cm³  and no. of unit cell present = 1.56×$10^{22}$