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How can you calculate the density of a cubic crystal whose length of the edge of the unit cell is known?

Answer»

Solution :Let the length of the edge of the cell = a pm The volume of the unit cell = `(a pm)^(3)`
`= ( a xx 10^(-10)CM)^(3)= a^(3) xx 10^(-30)cm^(3)`
`"Density of the unit cell "= ("Mass of unit cell")/("Volume of unit cell")`
`"Mass of unit cell" = "Number of atoms in unit cell" xx "Mass of each ATOM" = Z xx m`
where Z = number of atoms in unit cell and m = mass of each atom
`"Mass of each atom, m" = ("Atomic mass")/("Avogadro number")=(A)/(N_(0))`
`"Density" = ( Z xx A)/(a^(3) xx 10^(-30) xx N_(0))GCH^(-3)`
Density of unit cell is the same as the density of the substance.


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