1.

The cell edge of a fcc crystal is 100 pm and its density is 10.0 g cm^(-3) . The number of atoms in 100 g of this crystal is:

Answer»

`1 xx 10^(25)`
`2 xx 10^(25)`
` 3 xx 10^(25)`
`4 xx 10^(25)`

Solution :Volume of units cell `= ( 100 xx 10^(-10) cm)^(3)`
`= 10^(-24) cm^(3)`
Density `= 10.0 G cm^(-3)`
Mass of unit cell `= 10^(-24) xx 10`
`= 10 ^(-23) g `
No. of unit cells in 100 g `= ( 100)/( 10^(-23))= 10^(25)`
Since the lattice is fcc and each unit cell has 4 atoms PER unit cell.
No. of atoms `= 4 xx 10^(25)`


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