An element ' X' ( Atomic mass = 40 g / mol ) having f.c.c. structure,h

1.	An element ' X' ( Atomic mass = 40 g / mol ) having f.c.c. structure,has unit cell edge length of 400 pm.Calculate the density of 'X' and number of unit cell in 4gm of x
Answer» We know that : d= Z * M/a^3* NA Molar mass(M)=40 g /mol Edge length(a)=400 pm=40010^-10 cm Z for FCC=4 d=4 40 / [40010^-10]^3 6.02310^23 d= 4.1 g/cm^3 We know that, Number of atoms in a UNIT cell of FCC=4 Number of atoms in 4 g=4/40 X 6.02310^23 =6.02310^22 atoms . Since , 1 unit cell has------------>4 atoms ? < ------------------ 6.02310^22 atoms No. of unit cells=6.02310^22/4 =1.5010^22 unit cells in 4 g of X.

An element ' X' ( Atomic mass = 40 g / mol ) having f.c.c. structure,has unit cell edge length of 400 pm.Calculate the density of 'X' and number of unit cell in 4gm of x

Answer»

We know that :

d= Z * M/a^3* NA

Molar mass(M)=40 g /mol

Edge length(a)=400 pm=400*10^-10 cm

Z for FCC=4

d=4 * 40 / [400*10^-10]^3 * 6.023*10^23

d= 4.1 g/cm^3

We know that,

Number of atoms in a UNIT cell of FCC=4

Number of atoms in 4 g=4/40 X 6.023*10^23

=6.023*10^22 atoms .

Since , 1 unit cell has------------>4 atoms

? < ------------------ 6.023*10^22 atoms

No. of unit cells=6.023*10^22/4

=1.50*10^22 unit cells in 4 g of X.