1.

Total number of voids in 0.5 mole of a compound forming hexagonal closed packed structure are

Answer»

`60.22xx10^(23)`
`3.011xx10^(23)`
`9.034xx10^(23)`
`4.516xx10^(23)`

Solution :Number of CLOSE PACKED PARTICLES
`=0.5xx6.023xx10^(23)`
`:.` Number of octahedral voids `=3.0115xx10^(23)`
Number of tetrahedral voids `=2xx3.0115xx10^(23)`
`=6.023xx10^(23)`
THEREFORE total no of void `=6.023xx10^(23)+3.015xx10^(23)`
`=9.0345xx10^(23)`


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