Titanium metal crystallises in a BCC arrangement and radius of an atom is 142 picometre. Determine atomic weight of titanium if density is 16.6 g/cm^(3). Also determine the number of unit cells present in a 5 cm^(3)block of titanium metals.

Titanium metal crystallises in a BCC arrangement and radius of an atom

1.	Titanium metal crystallises in a BCC arrangement and radius of an atom is 142 picometre. Determine atomic weight of titanium if density is 16.6 g/cm^(3). Also determine the number of unit cells present in a 5 cm^(3)block of titanium metals.
Answer» Solution : In BCC arrangement of atoms the relatioship between edge length radius of atom is `4r=sqrt(3a)` Density `(rho)=(NM)/(N_(A)V)` Where, N = Number of atoms per unit cell, M = Molar mass, NA = Avogadro's number V = Volume of unit cell `implies16.6=(2M)/(6.023xx10^(23)((4)/(sqrt3)xx142xx10^(-10))^(3)` Here, radius is taken in cm unit since, density is in g/`cm^(3)` unit. Solving : M = 176. Also mass of 5 `cm^(3)` BLOCK `=5xx16.6`=83 g `implies` Number of atoms is 5 `cm^(3)` block `=(83)/(176)xx6.023xx10^(23)` `implies` Number of unit CELLS=`(No. of atoms)/(2)` `(83xx6.23xx10^(23))/(176xx2)=1.42xx10^(23)`