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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?
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Verified Answer
An element forms a cubic unit cell with edge length 405 pm. Molar mass...
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.
Most Upvoted Answer
An element forms a cubic unit cell with edge length 405 pm. Molar mass...
Given:
Edge length of cubic unit cell = 405 pm = 405 x 10^-12 m
Molar mass of the element = 2.7 x 10^-2 Kg/mol
Density of the element = 2.7 x 10^3 Kg/m^3
To find:
Number of atoms of the element present per unit cell.
Solution:
1. Calculation of volume of unit cell:
Volume of unit cell = (edge length)^3
= (405 x 10^-12 m)^3
= 6.67 x 10^-29 m^3
2. Calculation of mass of unit cell:
Mass of unit cell = Density x Volume
= 2.7 x 10^3 Kg/m^3 x 6.67 x 10^-29 m^3
= 1.8 x 10^-25 Kg
3. Calculation of number of moles of element in unit cell:
Number of moles = Mass / Molar mass
= 1.8 x 10^-25 Kg / 2.7 x 10^-2 Kg/mol
= 6.67 x 10^-24 mol
4. Calculation of number of atoms in unit cell:
As per the concept of cubic unit cell, there are 4 atoms present at the corners and 1 atom present at the center of the cube. Therefore, the total number of atoms present in the unit cell is 4 + 1 = 5.
5. Calculation of number of atoms per unit cell:
To find the number of atoms per unit cell, we need to divide the total number of atoms in the unit cell by Avogadro's number (6.022 x 10^23).
Number of atoms per unit cell = 5 / 6.022 x 10^23
= 8.31 x 10^-24
As this value is closest to 4, the correct answer is option 'A' i.e. 4 atoms.
Edge length of cubic unit cell = 405 pm = 405 x 10^-12 m
Molar mass of the element = 2.7 x 10^-2 Kg/mol
Density of the element = 2.7 x 10^3 Kg/m^3
To find:
Number of atoms of the element present per unit cell.
Solution:
1. Calculation of volume of unit cell:
Volume of unit cell = (edge length)^3
= (405 x 10^-12 m)^3
= 6.67 x 10^-29 m^3
2. Calculation of mass of unit cell:
Mass of unit cell = Density x Volume
= 2.7 x 10^3 Kg/m^3 x 6.67 x 10^-29 m^3
= 1.8 x 10^-25 Kg
3. Calculation of number of moles of element in unit cell:
Number of moles = Mass / Molar mass
= 1.8 x 10^-25 Kg / 2.7 x 10^-2 Kg/mol
= 6.67 x 10^-24 mol
4. Calculation of number of atoms in unit cell:
As per the concept of cubic unit cell, there are 4 atoms present at the corners and 1 atom present at the center of the cube. Therefore, the total number of atoms present in the unit cell is 4 + 1 = 5.
5. Calculation of number of atoms per unit cell:
To find the number of atoms per unit cell, we need to divide the total number of atoms in the unit cell by Avogadro's number (6.022 x 10^23).
Number of atoms per unit cell = 5 / 6.022 x 10^23
= 8.31 x 10^-24
As this value is closest to 4, the correct answer is option 'A' i.e. 4 atoms.
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Community Answer
An element forms a cubic unit cell with edge length 405 pm. Molar mass...
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
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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 10-2Kg/mol and its density is given as 2.7 X 103Kg/m3. How many atoms of this elements are present per unit cell.a)4b)2c)1d)6Correct answer is 'A'. Can you explain this answer?
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