An element crystallises in a b.c.c lattice with cell edge of 500 pm. The density of the element is 7.5g cm^{-3}. How many atoms are present in 300 g of the element?

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#### Solution

a = 500 pm = 500 x 10^{-10} cm

z = 2

m = 300 g

`Density(d)=(zM)/(a^2N_A)`

`7.5=(2xxm)/((500)^3xx10^(-30)xx6.02xx10^23)`

`M=(7.5xx(500)^3xx10(-30)xx6.02xx10^23)/2`

`M=282.18 " g/mol"`

`"Molar mass (M)"=("Mass of compound"xxN_A)/"Number of atoms"`

`282.18=(300xx6.02xx10^23)/"Number of atoms"`

Number of atoms = 6.4 x 10^{23 }

Concept: Number of Atoms in a Unit Cell

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