Advertisement Remove all ads

An Element Crystallizes in Fcc Lattice with a Cell Edge of 300 Pm. the Density of the Element is 10.8 G Cm−3. Calculate the Number of Atoms in 108 G of the Element. - Chemistry

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Numerical

An element crystallizes in fcc lattice with a cell edge of 300 pm. The density of the element is 10.8 g cm−3. Calculate the number of atoms in 108 g of the element.

Advertisement Remove all ads

Solution

Given: Mass of the element, W = 108 g

z (fcc) = 4  
Edge length, (a) = 300 pm    ...(1pm = 10−​10 cm)
Density, (​ρ) = 10.8 g/cm3

`"p" = (z xx "M")/(a^3 xx  "N"_"A")`

`10.8 = (4 xx "M")/((300 xx 10^-10)^3 xx6.022 xx 10^23)`

`"M" = (10.8 xx 2.7 xx 10^-23 xx 6.022 xx 10^23)/(4) "g"`

`n = "N"/"N"_("A") = "W"/"M"`

`"N" = "W"/"M" xx "N"_"A"`

`"N" = (108 xx 4 xx 6.022 xx 10^23)/(10.8 x 2.7 xx 10^-23 xx 6.022 xx 10^23)`

`"N" = 14.8 xx 10^23 "atoms"`

Concept: Calculations Involving Unit Cell Dimensions
  Is there an error in this question or solution?

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×