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A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm-3. The molar mass of the metal is (NA Avogadro's constant= 6.02 x 1023 mol-1)
- a)40 g mol-1
- b)30 g mol-1
- c)27 g mol-1
- d)20 g mol-1
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A metal has a fcc lattice. The edge length of the unit cell is 404 pm....
Given, cell is fcc, So Z =4
Edge length, a = 404 pm = 4.04 x 10-8 cm
Density of metal, d = 2.72 g cm-3
NA = 6.02 x 1023 mol-1
Molar mass ofg the metal, M =?
We know that
Edge length, a = 404 pm = 4.04 x 10-8 cm
Density of metal, d = 2.72 g cm-3
NA = 6.02 x 1023 mol-1
Molar mass ofg the metal, M =?
We know that
Most Upvoted Answer
A metal has a fcc lattice. The edge length of the unit cell is 404 pm....
Given Information
- A metal has a fcc lattice
- Edge length of the unit cell = 404 pm
- Density of the metal = 2.72 g cm-3
- Avogadro's number (NA) = 6.02 x 1023mol-1
To Find: Molar mass of the metal
Solution
1. Calculation of Volume of Unit Cell
- In an fcc lattice, there are 4 atoms per unit cell
- The distance between the opposite faces in a cube is equal to the edge length of the cube
- Therefore, the diagonal of a face of the cube is √2 times the edge length
- The diagonal of the cube is also the distance between the centers of two opposite corner atoms
- Hence, the length of the body diagonal of the cube is √3 times the edge length
- The volume of the unit cell can be calculated as follows:
Volume of Unit Cell = (Edge Length)^3 / 4
= (404 pm)^3 / 4
= 26.43 x 10^-24 cm^3
2. Calculation of Mass of One Unit Cell
- The density of the metal is given as 2.72 g cm^-3
- Therefore, the mass of the unit cell can be calculated as follows:
Mass of Unit Cell = Density x Volume of Unit Cell
= 2.72 g cm^-3 x 26.43 x 10^-24 cm^3
= 7.20 x 10^-23 g
3. Calculation of Molar Mass of the Metal
- The mass of one unit cell is equal to the mass of the atoms present in it
- The number of atoms present in one unit cell of an fcc lattice is 4
- Therefore, the mass of one atom can be calculated as follows:
Mass of one atom = mass of unit cell / number of atoms
= 7.20 x 10^-23 g / 4
= 1.80 x 10^-23 g
- The molar mass of the metal can be calculated as follows:
Molar Mass = Mass of one atom x Avogadro's number
= 1.80 x 10^-23 g x 6.02 x 1023 mol^-1
= 27.22 g mol^-1
Therefore, the molar mass of the metal is 27 g mol^-1, which is option 'C'.
- A metal has a fcc lattice
- Edge length of the unit cell = 404 pm
- Density of the metal = 2.72 g cm-3
- Avogadro's number (NA) = 6.02 x 1023mol-1
To Find: Molar mass of the metal
Solution
1. Calculation of Volume of Unit Cell
- In an fcc lattice, there are 4 atoms per unit cell
- The distance between the opposite faces in a cube is equal to the edge length of the cube
- Therefore, the diagonal of a face of the cube is √2 times the edge length
- The diagonal of the cube is also the distance between the centers of two opposite corner atoms
- Hence, the length of the body diagonal of the cube is √3 times the edge length
- The volume of the unit cell can be calculated as follows:
Volume of Unit Cell = (Edge Length)^3 / 4
= (404 pm)^3 / 4
= 26.43 x 10^-24 cm^3
2. Calculation of Mass of One Unit Cell
- The density of the metal is given as 2.72 g cm^-3
- Therefore, the mass of the unit cell can be calculated as follows:
Mass of Unit Cell = Density x Volume of Unit Cell
= 2.72 g cm^-3 x 26.43 x 10^-24 cm^3
= 7.20 x 10^-23 g
3. Calculation of Molar Mass of the Metal
- The mass of one unit cell is equal to the mass of the atoms present in it
- The number of atoms present in one unit cell of an fcc lattice is 4
- Therefore, the mass of one atom can be calculated as follows:
Mass of one atom = mass of unit cell / number of atoms
= 7.20 x 10^-23 g / 4
= 1.80 x 10^-23 g
- The molar mass of the metal can be calculated as follows:
Molar Mass = Mass of one atom x Avogadro's number
= 1.80 x 10^-23 g x 6.02 x 1023 mol^-1
= 27.22 g mol^-1
Therefore, the molar mass of the metal is 27 g mol^-1, which is option 'C'.
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Community Answer
A metal has a fcc lattice. The edge length of the unit cell is 404 pm....
Answer is c.. care the unit of edge length ,is cm
...d=zm÷a3.Na
...d=zm÷a3.Na
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A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm-3.The molar mass of the metal is (NAAvogadro's constant= 6.02 x 1023mol-1)a)40 g mol-1b)30 g mol-1c)27 g mol-1d)20 g mol-1Correct answer is option 'C'. Can you explain this answer?
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A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm-3.The molar mass of the metal is (NAAvogadro's constant= 6.02 x 1023mol-1)a)40 g mol-1b)30 g mol-1c)27 g mol-1d)20 g mol-1Correct answer is option 'C'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm-3.The molar mass of the metal is (NAAvogadro's constant= 6.02 x 1023mol-1)a)40 g mol-1b)30 g mol-1c)27 g mol-1d)20 g mol-1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm-3.The molar mass of the metal is (NAAvogadro's constant= 6.02 x 1023mol-1)a)40 g mol-1b)30 g mol-1c)27 g mol-1d)20 g mol-1Correct answer is option 'C'. Can you explain this answer?.
A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm-3.The molar mass of the metal is (NAAvogadro's constant= 6.02 x 1023mol-1)a)40 g mol-1b)30 g mol-1c)27 g mol-1d)20 g mol-1Correct answer is option 'C'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm-3.The molar mass of the metal is (NAAvogadro's constant= 6.02 x 1023mol-1)a)40 g mol-1b)30 g mol-1c)27 g mol-1d)20 g mol-1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm-3.The molar mass of the metal is (NAAvogadro's constant= 6.02 x 1023mol-1)a)40 g mol-1b)30 g mol-1c)27 g mol-1d)20 g mol-1Correct answer is option 'C'. Can you explain this answer?.
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