1.

An element crystallizes in a fee lattice with cell edge of 400 pm. The density of the element is 7 g//cm3. How many atoms one present in 280 g of the element ?

Answer»

SOLUTION :`"Volume of unit cell"=a^(3)"(a = edge length)"`
`="400 pm"`
`=(400xx10^(-12)m)^(3)`
`=(400xx10^(-10)cm)^(3)=64xx10^(-24)cm^(3)`
`"Volume of 208 g of the element"=(208)/(7g//cm^(3))=29.71cm^(3)`
`"Number of unit cells in this volume"=("VOL. of given amount")/("Vol of ONE unit cell")`
`=(29.71)/(64xx10^(-24))=0.46xx10^(24)`
Since each f.c.c. unit cell containss 4 atoms therefore,
`"Total number "=4xx0.46xx1024`
`=1.84xx"1024 atoms."`


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