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Question
Silver crystallises in FCC structure. If density of silver is 10.51 gcm-3, calculate the volume of unit cell. [Atomic mass of slive (Ag) = 108 gm-1]
Solution
Density of Ag = 10.51 g/cm3
Vol. of unit cell = ?
Mass of one atom of silver
`="molar mass of silver"/"N"_"A"`
`= 108/(6.022 xx 10^23)`
`= 17.93 xx 10^-23`
Mass of unit cell of silver = Atoms in unit cell × Mass of 1 atom
`= 4 xx 17.93 xx 10^-23`
`= 71.72 xx 10^-23`
`therefore "Density of Ag" = "mass of unit cell"/"Vol.of unit cell"`
`therefore "Vol of unit cell" = "mass of unit cell"/"density of Ag"`
`= (71.72 xx 10^-23)/10.51`
col.of.unit cell = 68.24 × 10-23 cm3
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