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Lithium forms a BCC lattice with an edge length of 350 pm. The experimental density of lithium is 0.53 g cm-3. What is the percentage of missing lithium atoms? (Atomic mass of Lithium = 7 amu)
- a)97.7%
- b)95.4%
- c)4.6%
- d)2.3%
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Lithium forms a BCC lattice with an edge length of 350 pm. The experim...
Given,
Edge length (a) = 350 pm = 3.5 x 10-8 cm
Atomic mass (M) = 7 amu
Avogadro’s number (N0) = 6.02 x 1023
Density (ρ) = (Z x M)/(a3 x N0)
= (2 x 7)/((3.5 x 10-8)3 x 6.02 x 1023)
= 0.542 g cm-3
% of lithium atoms occupied = (Experimental density/Theoretical density) x 100
= 0.53/0.542 x 100
= 97.7%
% of unoccupied lattice sites = 100 – 97.7
= 2.3%.
Edge length (a) = 350 pm = 3.5 x 10-8 cm
Atomic mass (M) = 7 amu
Avogadro’s number (N0) = 6.02 x 1023
Density (ρ) = (Z x M)/(a3 x N0)
= (2 x 7)/((3.5 x 10-8)3 x 6.02 x 1023)
= 0.542 g cm-3
% of lithium atoms occupied = (Experimental density/Theoretical density) x 100
= 0.53/0.542 x 100
= 97.7%
% of unoccupied lattice sites = 100 – 97.7
= 2.3%.
Most Upvoted Answer
Lithium forms a BCC lattice with an edge length of 350 pm. The experim...
Given:
- Lithium forms a BCC lattice with an edge length of 350 pm.
- Experimental density of lithium is 0.53 g cm-3.
- Atomic mass of Lithium = 7 amu.
To Find:
- The percentage of missing lithium atoms.
Concepts Used:
- BCC lattice structure.
- Calculation of the number of atoms in a BCC unit cell.
- Calculation of the density of a substance.
- Calculation of the percentage of missing atoms.
Explanation:
Step 1: Calculate the number of atoms in a BCC unit cell:
- In a BCC lattice, there are 2 atoms per unit cell.
- Each corner atom is shared by 8 unit cells, and each unit cell contributes 1/8th of an atom.
- There is 1 atom at the center of the unit cell, which is not shared.
- Therefore, the total number of atoms in a BCC unit cell is (1/8) * 8 + 1 = 2.
Step 2: Calculate the volume of the unit cell:
- The edge length of the BCC unit cell is given as 350 pm.
- The volume of a cube is calculated by multiplying the edge length by itself three times.
- Therefore, the volume of the unit cell is (350 pm)^3.
Step 3: Calculate the mass of the unit cell:
- The density of lithium is given as 0.53 g cm-3.
- Density is defined as mass divided by volume.
- Therefore, the mass of the unit cell is the product of the density and the volume of the unit cell.
Step 4: Calculate the molar mass of the unit cell:
- The atomic mass of lithium is given as 7 amu.
- Molar mass is defined as the mass of one mole of a substance.
- Therefore, the molar mass of the unit cell is the product of the atomic mass of lithium and the number of atoms in the unit cell.
Step 5: Calculate the number of moles of lithium in the unit cell:
- The number of moles is calculated by dividing the mass of the unit cell by the molar mass of the unit cell.
Step 6: Calculate the number of missing lithium atoms:
- The number of missing lithium atoms is calculated by subtracting the number of moles of lithium in the unit cell from the number of atoms in the unit cell.
Step 7: Calculate the percentage of missing lithium atoms:
- The percentage of missing lithium atoms is calculated by dividing the number of missing lithium atoms by the total number of atoms in the unit cell and multiplying by 100.
Answer:
The percentage of missing lithium atoms is 2.3%.
- Lithium forms a BCC lattice with an edge length of 350 pm.
- Experimental density of lithium is 0.53 g cm-3.
- Atomic mass of Lithium = 7 amu.
To Find:
- The percentage of missing lithium atoms.
Concepts Used:
- BCC lattice structure.
- Calculation of the number of atoms in a BCC unit cell.
- Calculation of the density of a substance.
- Calculation of the percentage of missing atoms.
Explanation:
Step 1: Calculate the number of atoms in a BCC unit cell:
- In a BCC lattice, there are 2 atoms per unit cell.
- Each corner atom is shared by 8 unit cells, and each unit cell contributes 1/8th of an atom.
- There is 1 atom at the center of the unit cell, which is not shared.
- Therefore, the total number of atoms in a BCC unit cell is (1/8) * 8 + 1 = 2.
Step 2: Calculate the volume of the unit cell:
- The edge length of the BCC unit cell is given as 350 pm.
- The volume of a cube is calculated by multiplying the edge length by itself three times.
- Therefore, the volume of the unit cell is (350 pm)^3.
Step 3: Calculate the mass of the unit cell:
- The density of lithium is given as 0.53 g cm-3.
- Density is defined as mass divided by volume.
- Therefore, the mass of the unit cell is the product of the density and the volume of the unit cell.
Step 4: Calculate the molar mass of the unit cell:
- The atomic mass of lithium is given as 7 amu.
- Molar mass is defined as the mass of one mole of a substance.
- Therefore, the molar mass of the unit cell is the product of the atomic mass of lithium and the number of atoms in the unit cell.
Step 5: Calculate the number of moles of lithium in the unit cell:
- The number of moles is calculated by dividing the mass of the unit cell by the molar mass of the unit cell.
Step 6: Calculate the number of missing lithium atoms:
- The number of missing lithium atoms is calculated by subtracting the number of moles of lithium in the unit cell from the number of atoms in the unit cell.
Step 7: Calculate the percentage of missing lithium atoms:
- The percentage of missing lithium atoms is calculated by dividing the number of missing lithium atoms by the total number of atoms in the unit cell and multiplying by 100.
Answer:
The percentage of missing lithium atoms is 2.3%.
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Community Answer
Lithium forms a BCC lattice with an edge length of 350 pm. The experim...
Given,
Edge length (a) = 350 pm = 3.5 x 10-8 cm
Atomic mass (M) = 7 amu
Avogadro’s number (N0) = 6.02 x 1023
Density (ρ) = (Z x M)/(a3 x N0)
= (2 x 7)/((3.5 x 10-8)3 x 6.02 x 1023)
= 0.542 g cm-3
% of lithium atoms occupied = (Experimental density/Theoretical density) x 100
= 0.53/0.542 x 100
= 97.7%
% of unoccupied lattice sites = 100 – 97.7
= 2.3%.
Edge length (a) = 350 pm = 3.5 x 10-8 cm
Atomic mass (M) = 7 amu
Avogadro’s number (N0) = 6.02 x 1023
Density (ρ) = (Z x M)/(a3 x N0)
= (2 x 7)/((3.5 x 10-8)3 x 6.02 x 1023)
= 0.542 g cm-3
% of lithium atoms occupied = (Experimental density/Theoretical density) x 100
= 0.53/0.542 x 100
= 97.7%
% of unoccupied lattice sites = 100 – 97.7
= 2.3%.
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Lithium forms a BCC lattice with an edge length of 350 pm. The experimental density of lithium is 0.53 g cm-3. What is the percentage of missing lithium atoms? (Atomic mass of Lithium = 7 amu)a)97.7%b)95.4%c)4.6%d)2.3%Correct answer is option 'D'. Can you explain this answer?
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Lithium forms a BCC lattice with an edge length of 350 pm. The experimental density of lithium is 0.53 g cm-3. What is the percentage of missing lithium atoms? (Atomic mass of Lithium = 7 amu)a)97.7%b)95.4%c)4.6%d)2.3%Correct answer is option 'D'. 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 Lithium forms a BCC lattice with an edge length of 350 pm. The experimental density of lithium is 0.53 g cm-3. What is the percentage of missing lithium atoms? (Atomic mass of Lithium = 7 amu)a)97.7%b)95.4%c)4.6%d)2.3%Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Lithium forms a BCC lattice with an edge length of 350 pm. The experimental density of lithium is 0.53 g cm-3. What is the percentage of missing lithium atoms? (Atomic mass of Lithium = 7 amu)a)97.7%b)95.4%c)4.6%d)2.3%Correct answer is option 'D'. Can you explain this answer?.
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Solutions for Lithium forms a BCC lattice with an edge length of 350 pm. The experimental density of lithium is 0.53 g cm-3. What is the percentage of missing lithium atoms? (Atomic mass of Lithium = 7 amu)a)97.7%b)95.4%c)4.6%d)2.3%Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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Lithium forms a BCC lattice with an edge length of 350 pm. The experimental density of lithium is 0.53 g cm-3. What is the percentage of missing lithium atoms? (Atomic mass of Lithium = 7 amu)a)97.7%b)95.4%c)4.6%d)2.3%Correct answer is option 'D'. Can you explain this answer?, a detailed solution for Lithium forms a BCC lattice with an edge length of 350 pm. The experimental density of lithium is 0.53 g cm-3. What is the percentage of missing lithium atoms? (Atomic mass of Lithium = 7 amu)a)97.7%b)95.4%c)4.6%d)2.3%Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of Lithium forms a BCC lattice with an edge length of 350 pm. The experimental density of lithium is 0.53 g cm-3. What is the percentage of missing lithium atoms? (Atomic mass of Lithium = 7 amu)a)97.7%b)95.4%c)4.6%d)2.3%Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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