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An element with cell edge of 288 pm has a density of 7.2 g cm-3. What type of structure does the element have if it’s atomic mass M=51.8 g mol-1?
- a)Body-Centred Cubic (BCC)
- b)Face-Centred Cubic (FCC)
- c)Simple Cubic
- d)Hexagonal Closed Packing (HCP)
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
An element with cell edge of 288 pm has a density of 7.2 g cm-3. What ...
Given,
Edge length (a) = 288 pm
Density (ρ) = 7.2 g cm-3
Atomic mass (M) = 51.8 g mol-1
Avogadro’s number (N0) = 6.02 x 1023
We know, (ρ) = (Z x M)/(a3 x N0)
Or Z =(ρ x a3 x N0)/M = (7.2 x (288 x 10-10) 3 x 6.02 x 1023)/51.8
Z = 2
Therefore, the element has Body-Centred Cubic (BCC) type of structure.
Edge length (a) = 288 pm
Density (ρ) = 7.2 g cm-3
Atomic mass (M) = 51.8 g mol-1
Avogadro’s number (N0) = 6.02 x 1023
We know, (ρ) = (Z x M)/(a3 x N0)
Or Z =(ρ x a3 x N0)/M = (7.2 x (288 x 10-10) 3 x 6.02 x 1023)/51.8
Z = 2
Therefore, the element has Body-Centred Cubic (BCC) type of structure.
Most Upvoted Answer
An element with cell edge of 288 pm has a density of 7.2 g cm-3. What ...
Given,
Edge length (a) = 288 pm
Density (ρ) = 7.2 g cm-3
Atomic mass (M) = 51.8 g mol-1
Avogadro’s number (N0) = 6.02 x 1023
We know, (ρ) = (Z x M)/(a3 x N0)
Or Z =(ρ x a3 x N0)/M = (7.2 x (288 x 10-10) 3 x 6.02 x 1023)/51.8
Z = 2
Therefore, the element has Body-Centred Cubic (BCC) type of structure.
Edge length (a) = 288 pm
Density (ρ) = 7.2 g cm-3
Atomic mass (M) = 51.8 g mol-1
Avogadro’s number (N0) = 6.02 x 1023
We know, (ρ) = (Z x M)/(a3 x N0)
Or Z =(ρ x a3 x N0)/M = (7.2 x (288 x 10-10) 3 x 6.02 x 1023)/51.8
Z = 2
Therefore, the element has Body-Centred Cubic (BCC) type of structure.
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Community Answer
An element with cell edge of 288 pm has a density of 7.2 g cm-3. What ...
The type of structure an element has can be determined by its cell edge length and density.
To determine the type of structure, we need to calculate the number of atoms per unit cell.
The formula to calculate the number of atoms per unit cell is:
Number of atoms per unit cell = (Density * Avogadro's number) / (Atomic mass * Volume of unit cell)
Given:
Cell edge length = 288 pm = 288 x 10^-12 m
Density = 7.2 g cm^-3 = 7.2 x 10^3 kg m^-3
We need to convert the density to kg m^-3 to match the units of the cell edge length.
To convert g cm^-3 to kg m^-3:
1 g cm^-3 = 1 x 10^3 kg m^-3
Therefore, the density in kg m^-3 is 7.2 x 10^3 kg m^-3.
Now, we need to calculate the volume of the unit cell.
The volume of the unit cell can be calculated using the formula:
Volume of unit cell = (Cell edge length)^3
Volume of unit cell = (288 x 10^-12 m)^3 = 2.985984 x 10^-24 m^3
Now, we can calculate the number of atoms per unit cell:
Number of atoms per unit cell = (Density * Avogadro's number) / (Atomic mass * Volume of unit cell)
Number of atoms per unit cell = (7.2 x 10^3 kg m^-3 * 6.022 x 10^23 mol^-1) / (Atomic mass * 2.985984 x 10^-24 m^3)
Since we don't have the atomic mass of the element, we cannot calculate the number of atoms per unit cell or determine the type of structure.
To determine the type of structure, we need to calculate the number of atoms per unit cell.
The formula to calculate the number of atoms per unit cell is:
Number of atoms per unit cell = (Density * Avogadro's number) / (Atomic mass * Volume of unit cell)
Given:
Cell edge length = 288 pm = 288 x 10^-12 m
Density = 7.2 g cm^-3 = 7.2 x 10^3 kg m^-3
We need to convert the density to kg m^-3 to match the units of the cell edge length.
To convert g cm^-3 to kg m^-3:
1 g cm^-3 = 1 x 10^3 kg m^-3
Therefore, the density in kg m^-3 is 7.2 x 10^3 kg m^-3.
Now, we need to calculate the volume of the unit cell.
The volume of the unit cell can be calculated using the formula:
Volume of unit cell = (Cell edge length)^3
Volume of unit cell = (288 x 10^-12 m)^3 = 2.985984 x 10^-24 m^3
Now, we can calculate the number of atoms per unit cell:
Number of atoms per unit cell = (Density * Avogadro's number) / (Atomic mass * Volume of unit cell)
Number of atoms per unit cell = (7.2 x 10^3 kg m^-3 * 6.022 x 10^23 mol^-1) / (Atomic mass * 2.985984 x 10^-24 m^3)
Since we don't have the atomic mass of the element, we cannot calculate the number of atoms per unit cell or determine the type of structure.
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An element with cell edge of 288 pm has a density of 7.2 g cm-3. What type of structure does the element have if it’s atomic mass M=51.8 g mol-1?a)Body-Centred Cubic (BCC)b)Face-Centred Cubic (FCC)c)Simple Cubicd)Hexagonal Closed Packing (HCP)Correct answer is option 'A'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about An element with cell edge of 288 pm has a density of 7.2 g cm-3. What type of structure does the element have if it’s atomic mass M=51.8 g mol-1?a)Body-Centred Cubic (BCC)b)Face-Centred Cubic (FCC)c)Simple Cubicd)Hexagonal Closed Packing (HCP)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An element with cell edge of 288 pm has a density of 7.2 g cm-3. What type of structure does the element have if it’s atomic mass M=51.8 g mol-1?a)Body-Centred Cubic (BCC)b)Face-Centred Cubic (FCC)c)Simple Cubicd)Hexagonal Closed Packing (HCP)Correct answer is option 'A'. Can you explain this answer?.
An element with cell edge of 288 pm has a density of 7.2 g cm-3. What type of structure does the element have if it’s atomic mass M=51.8 g mol-1?a)Body-Centred Cubic (BCC)b)Face-Centred Cubic (FCC)c)Simple Cubicd)Hexagonal Closed Packing (HCP)Correct answer is option 'A'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about An element with cell edge of 288 pm has a density of 7.2 g cm-3. What type of structure does the element have if it’s atomic mass M=51.8 g mol-1?a)Body-Centred Cubic (BCC)b)Face-Centred Cubic (FCC)c)Simple Cubicd)Hexagonal Closed Packing (HCP)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An element with cell edge of 288 pm has a density of 7.2 g cm-3. What type of structure does the element have if it’s atomic mass M=51.8 g mol-1?a)Body-Centred Cubic (BCC)b)Face-Centred Cubic (FCC)c)Simple Cubicd)Hexagonal Closed Packing (HCP)Correct answer is option 'A'. Can you explain this answer?.
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