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An element forms a cubic unit cell with edge length 405 pm. Molar mass of this element is 2.7 X 10-2 Kg/mol and its density is given as 2.7 X 103 Kg/m3. How many atoms of this elements are present per unit cell.
  • a)
    4
  • b)
    2
  • c)
    1
  • d)
    6
Correct answer is 'A'. Can you explain this answer?

Class 12 Question

Volume of the cube =(405×10^-12)^3 m^3=6.64×10^-29 m^3 net weight of the element= density × volume=1.79× 10^25 kg mole number= net weight/ molar mass= 6.62 ×10^-24 mol atom number= mole number× Avogadro's number= 3.99 (approx)= 4

Neeraj Chauhan
May 25, 2018
It is given that density of the element, d = 2.7 x 10^3 kg m-3

Molar mass, M = 2.7 x 10^-2 kg mol^-1

Edge length, a= 405 pm = 405 x 10^-12 m = 4.05 x 10^-10 m

It is known that, Avogadro's number, NA= 6.022 x 10^23 mol^-1

d = (z.M)/(a^3.NA)
z = (d.a^3 NA)/M
  = [2.7 x 10^3 kg m^-3 x (4.05 x 10^-10 m)^3 x 6.022 x 10^23 mol^-1]/2.7 x 10^-2 kg mol^-1

  = 4.004
  = 4
Applying the relation, This implies that four atoms of the element are present per unit cell. 

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