1.

The density and edge length values for a crystalline element with fcc lattice are 10 g cm^(-3) and 400 pm respectively. The number of unit cells in 32 g of this crystal is

Answer»

`8xx10^23`
`5xx10^22`
`8xx10^22`
`5xx10^23`

SOLUTION :Volume of the unit cell=`a^3="(400 pm)"^3`
`=(400xx10^(-10)cm)^3 =64 XX 10^(-24)cm^3`
Density of the unit cell=`10g//cm^3`
Mass of unit cell=Volumex density
`=64xx10^(-24)xx10=640xx10^(-24) g`
Now, `640xx10^(-24) g -=` 1 unit cell
`therefore 32 g -= 1/(640xx10^(-24))xx32=0.05xx10^24=5xx10^22`


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