Question
Sat July 02, 2011

# cesium chlorie crystallises as a body centredd cubic lattice has a density of 4.0g/cm^3 . calculate the edge lehgth of the unit cell of cesium chloride crystal

Mon July 04, 2011
CsCl crystallises as body centered cubic =4
i.e. Chloride ions are present at the corners=8 Chlorine atoms at 8 corners and the contribution of the atoms at the centrer=1/8.
Therefore, the atoms per unit cell=1 and 1 Cspresent at the centre.
Therefore, no. of formla units of CsCl present per unit cell=1,i.e.Z=1
Now density=mass/volume.
Volume =a3.
mass of CsCl=(132.9+35.5)
Mass of 1 molecule of CsCl=1* (132.9+35.5)/(6.023*1023)

Therefore,Volume=mass/density

a3= 27.9595*10-23/4.0=0.069898*10-21
a= (0.069898*10-21)1/3
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