1.

An atom crystallizes in fee erystal lattice and has a density of "10 gcm"^(-3) with unit cell edge length of 100pm. calcutate the number of atoms present in 1 g of crystal.

Answer»

Solution :`"Given,Density"="10 g cm"^(-3)`
mass = 1 g
`"Edge length of unit cell"="100 pm"`
`"Volume"=("mass")/("density")=(1g)/("10 g cm"^(-3))`
`=0.1cm^(3)`
`"Volume of unit cell"=a^(3)`
`=(100xx10^(-10)" cm")^(3)`
`=1xx10^(-24)cm^(3)`
Number of unit cell in 1 g of CRYSTAL,
`=("TOTAL volume")/("Volume of unit cell")`
`=(0.1cm^(3))/(1xx10^(-24)cm^(3))`
The given unit cell is of FCC type. Therefore, it contains 4 atoms.
`0.1xx10^(24)` unit cells will contain `4xx0.1xx10^(24)=4xx10^(23)` atoms


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