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Question
The number of atom in 100g of an FCC crystal with density d=10g/cm3 and cell edge equal to 100 pm is equal to
The number of atom in 100g of an FCC crystal with density d=10g/cm3 and cell edge equal to 100 pm is equal to
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Solution
d = Z x M/a3 x NA
where, a is edge of unit cell
NA = Avogadro number (6.022 × 1023)
M = Molar mass
Z = number of atoms per unit cell
For fcc, Z = 4
From the given data we can calculate the Molar mass of given substance as:
M = d x a3 x NA /z
Number of atoms = givan mass / molar mass(M) x NA
Putting the vaue of M in above equation we get
Number of moles = given mass x z x NA / d x a3 x NA
Number of atoms = given mass x z / d x a3
putting the values we get:
Number of atoms = ( 100 g x 4) /( 10 gcm^-3 x (100 x 10-10 cm)^3 )= 400 / 10 x 10^-24 = 4 x 10^25 atoms
where, a is edge of unit cell
NA = Avogadro number (6.022 × 1023)
M = Molar mass
Z = number of atoms per unit cell
For fcc, Z = 4
From the given data we can calculate the Molar mass of given substance as:
M = d x a3 x NA /z
Number of atoms = givan mass / molar mass(M) x NA
Putting the vaue of M in above equation we get
Number of moles = given mass x z x NA / d x a3 x NA
Number of atoms = given mass x z / d x a3
putting the values we get:
Number of atoms = ( 100 g x 4) /( 10 gcm^-3 x (100 x 10-10 cm)^3 )= 400 / 10 x 10^-24 = 4 x 10^25 atoms
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