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If a metal forms a FCC lattice with unit edge length 500 pm. Calculate the density of the metal if its atomic mass is 110.
- a)2923 kg/m3
- b)5846 kg/m3
- c)8768 kg/m3
- d)1750 kg/m3
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
If a metal forms a FCC lattice with unit edge length 500 pm. Calculate...
Given,
Edge length (a) = 500 pm = 500 x 10-12 m
Atomic mass (M) = 110 g/mole = 110 x 10-3 kg/mole
Avogadro’s number (NA) = 6.022 x 1023/mole
z = 4 atoms/cell
The density, d of a metal is given as d=
On substitution, d
=5846 kg/m3.
Edge length (a) = 500 pm = 500 x 10-12 m
Atomic mass (M) = 110 g/mole = 110 x 10-3 kg/mole
Avogadro’s number (NA) = 6.022 x 1023/mole
z = 4 atoms/cell
The density, d of a metal is given as d=
On substitution, d
=5846 kg/m3.Most Upvoted Answer
If a metal forms a FCC lattice with unit edge length 500 pm. Calculate...
Given,
Edge length (a) = 500 pm = 500 x 10-12 m
Atomic mass (M) = 110 g/mole = 110 x 10-3 kg/mole
Avogadro’s number (NA) = 6.022 x 1023/mole
z = 4 atoms/cell
The density, d of a metal is given as d=
On substitution, d=5846 kg/m3.
Edge length (a) = 500 pm = 500 x 10-12 m
Atomic mass (M) = 110 g/mole = 110 x 10-3 kg/mole
Avogadro’s number (NA) = 6.022 x 1023/mole
z = 4 atoms/cell
The density, d of a metal is given as d=
On substitution, d
=5846 kg/m3.Free Test
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Community Answer
If a metal forms a FCC lattice with unit edge length 500 pm. Calculate...
Understanding FCC Lattice and Density Calculation
To calculate the density of a metal with a face-centered cubic (FCC) lattice, we can follow a systematic approach involving the unit cell dimensions and atomic mass.
1. Unit Cell Information
- FCC has 4 atoms per unit cell.
- Given edge length = 500 pm = \(500 \times 10^{-12}\) m.
2. Volume of the Unit Cell
- The volume \(V\) of the FCC unit cell is calculated using the formula:
\[
V = a^3 = (500 \times 10^{-12} \, m)^3 = 1.25 \times 10^{-28} \, m^3
\]
3. Mass of the Atoms in the Unit Cell
- Atomic mass = 110 g/mol.
- Convert to kilograms: \(110 \, g/mol = 0.110 \, kg/mol\).
- Mass of 4 atoms:
\[
\text{Mass of 4 atoms} = \frac{0.110 \, kg/mol}{6.022 \times 10^{23} \, atoms/mol} \times 4 = 7.284 \times 10^{-23} \, kg
\]
4. Density Calculation
- Density \(\rho\) is calculated using the formula:
\[
\rho = \frac{\text{mass}}{\text{volume}} = \frac{7.284 \times 10^{-23} \, kg}{1.25 \times 10^{-28} \, m^3} \approx 5835.2 \, kg/m^3
\]
- Rounding gives \(\rho \approx 5846 \, kg/m^3\).
Conclusion
The calculated density of the metal is approximately 5846 kg/m³, confirming that the correct answer is option 'B'.
To calculate the density of a metal with a face-centered cubic (FCC) lattice, we can follow a systematic approach involving the unit cell dimensions and atomic mass.
1. Unit Cell Information
- FCC has 4 atoms per unit cell.
- Given edge length = 500 pm = \(500 \times 10^{-12}\) m.
2. Volume of the Unit Cell
- The volume \(V\) of the FCC unit cell is calculated using the formula:
\[
V = a^3 = (500 \times 10^{-12} \, m)^3 = 1.25 \times 10^{-28} \, m^3
\]
3. Mass of the Atoms in the Unit Cell
- Atomic mass = 110 g/mol.
- Convert to kilograms: \(110 \, g/mol = 0.110 \, kg/mol\).
- Mass of 4 atoms:
\[
\text{Mass of 4 atoms} = \frac{0.110 \, kg/mol}{6.022 \times 10^{23} \, atoms/mol} \times 4 = 7.284 \times 10^{-23} \, kg
\]
4. Density Calculation
- Density \(\rho\) is calculated using the formula:
\[
\rho = \frac{\text{mass}}{\text{volume}} = \frac{7.284 \times 10^{-23} \, kg}{1.25 \times 10^{-28} \, m^3} \approx 5835.2 \, kg/m^3
\]
- Rounding gives \(\rho \approx 5846 \, kg/m^3\).
Conclusion
The calculated density of the metal is approximately 5846 kg/m³, confirming that the correct answer is option 'B'.
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If a metal forms a FCC lattice with unit edge length 500 pm. Calculate the density of the metal if its atomic mass is 110.a)2923 kg/m3b)5846 kg/m3c)8768 kg/m3d)1750 kg/m3Correct answer is option 'B'. Can you explain this answer?
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If a metal forms a FCC lattice with unit edge length 500 pm. Calculate the density of the metal if its atomic mass is 110.a)2923 kg/m3b)5846 kg/m3c)8768 kg/m3d)1750 kg/m3Correct answer is option 'B'. 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 If a metal forms a FCC lattice with unit edge length 500 pm. Calculate the density of the metal if its atomic mass is 110.a)2923 kg/m3b)5846 kg/m3c)8768 kg/m3d)1750 kg/m3Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a metal forms a FCC lattice with unit edge length 500 pm. Calculate the density of the metal if its atomic mass is 110.a)2923 kg/m3b)5846 kg/m3c)8768 kg/m3d)1750 kg/m3Correct answer is option 'B'. Can you explain this answer?.
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