Question

Chromium crystallises in BCC structure if its Atomic diameter is 245pm.Find the density and atomic mass of Chromium 52 amu. Avogadro number: 6.022*1023

Answer 1

Answer : Density = 7.628 $g/Cm^{3}$

Solution : Given,

Number of atom in unit cell of BCC (Z) = 2

Diameter (d) = 245 pm

Atomic mass of chromium (M) = 52 amu

Avogadro's number $(N_{A})$ = 6.022 × $10^{23} mol^{-1}$

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$                  .........(1)

where,

$\rho$ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number

a = edge length of unit cell

Now, first we calculate the edge length of unit cell (a).

Formula used : The realtion between r, d and a shown in attached image. So, we get

$a=\frac{2\times d}{\sqrt{3}}$

put all the values in this formula, we get

d = 245 pm = 2.45 × $10^{-8}$ Cm          

conversion :   1 pm = 1 × $10^{-10}$ Cm

a = $\frac{2\times 2.45\times 10^{-8} }{\sqrt{3}}$ = 2.829 × $10^{-8}$ Cm

Volume of unit cell = $a^{3}$ = $(2.829\times 10^{-8} )^{3}$ = 22.64 × $10^{-24}Cm^{3}$

Now put all the values in above formula (1), we get

$\rho=\frac{2\times (52g/mol)}{(6.022\times 10^{23}mol^{-1}) \times(22.4\times 10^{-24}Cm^{3})}$ = 7.628 $g/Cm^{3}$