The compound CuCl has ZnS structure and edge length of unit cell is 500 pm.Calculate its density.(Atomic mass of Cu=63,Cl=35.5)
Answers
Answered by
44
Since ZnS has a face-centered cubic structure, it contains 4 sphere
so Z=(6x(1/2))+(8x(1/8)) = 4.
Density id given by = mass/volume
mass=zxm
volume=Na x a³
So, density=(z x m)÷(Na x a³)
where
a=500 pm = 5x10^-8 cm
M = 63+35.5 = 98.5
Na= 6.023×10^23
the density of CuCl is as follows
=[{(63+35.5)×4}÷{(6.023×10^23)×(5×10^-8)³}] g/cc
=[394÷(752.875x10^-1)]
=0.523327x10 g/cc
=5.23 g/cc
If u still have any doubt please ask in comment section or just DM
so Z=(6x(1/2))+(8x(1/8)) = 4.
Density id given by = mass/volume
mass=zxm
volume=Na x a³
So, density=(z x m)÷(Na x a³)
where
a=500 pm = 5x10^-8 cm
M = 63+35.5 = 98.5
Na= 6.023×10^23
the density of CuCl is as follows
=[{(63+35.5)×4}÷{(6.023×10^23)×(5×10^-8)³}] g/cc
=[394÷(752.875x10^-1)]
=0.523327x10 g/cc
=5.23 g/cc
If u still have any doubt please ask in comment section or just DM
Answered by
13
Heya..
Here is your answer...
ZnS has a FCC structure , so the value of Z = 4.
Molar mass of CuCl = 63.5 + 35.5 = 99 g
Density = 3.4 g cm-3
It may help you...☺☺
Here is your answer...
ZnS has a FCC structure , so the value of Z = 4.
Molar mass of CuCl = 63.5 + 35.5 = 99 g
Density = 3.4 g cm-3
It may help you...☺☺
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