# 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? - Chemistry

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?

#### 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 1023

Concept: Number of Atoms in a Unit Cell
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