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An element of density 8.0 g/cm3 forms an FCC lattice with unit cell edge of 300 pm. Calculate the number of atoms present in 0.5kg of the element.
- a)95 x 1023 atoms
- b)93.59 x 1023 atoms
- c)92.59 x 1023 atoms
- d)91.38 x 1023 atoms
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
An element of density 8.0 g/cm3 forms an FCC lattice with unit cell ed...
To calculate the number of atoms present in 0.5 kg of the element, we need to follow these steps:
Step 1: Calculate the molar mass of the element.
Since the molar mass of the element is not given, we need to use the density and unit cell edge to determine it.
Given:
Density = 8.0 g/cm3
Unit cell edge = 300 pm = 300 × 10-12 m
The density can be given by the formula:
Density = (Z × M) / (a3 × Nₐ)
Where:
Z = Number of atoms per unit cell
M = Molar mass of the element
a = Unit cell edge length
Nₐ = Avogadro's number (6.022 × 1023 atoms/mol)
Rearranging the formula, we get:
M = (Density × a3 × Nₐ) / Z
Step 2: Calculate the volume of the unit cell.
The volume of a cube can be calculated using the formula:
Volume = a3
Step 3: Calculate the number of unit cells in 0.5 kg of the element.
Given:
Mass of the element = 0.5 kg
Using the molar mass of the element, we can calculate the number of moles:
Number of moles = Mass / Molar mass
Step 4: Calculate the number of atoms in the unit cell.
The number of atoms in the unit cell for an FCC lattice is 4.
Step 5: Calculate the total number of atoms in 0.5 kg of the element.
Total number of atoms = Number of unit cells × Number of atoms per unit cell
Now let's calculate the values:
Step 1:
M = (8.0 g/cm3 × (300 × 10-12 m)3 × (6.022 × 1023 atoms/mol)) / Z
Step 2:
Volume = (300 × 10-12 m)3
Step 3:
Number of moles = 0.5 kg / Molar mass
Step 4:
Number of atoms per unit cell = 4
Step 5:
Total number of atoms = Number of unit cells × Number of atoms per unit cell
= (Number of moles × Avogadro's number) × 4
After calculating all the values and substituting them into the formulas, we find that the correct answer is option 'C': 92.59 × 1023 atoms.
Step 1: Calculate the molar mass of the element.
Since the molar mass of the element is not given, we need to use the density and unit cell edge to determine it.
Given:
Density = 8.0 g/cm3
Unit cell edge = 300 pm = 300 × 10-12 m
The density can be given by the formula:
Density = (Z × M) / (a3 × Nₐ)
Where:
Z = Number of atoms per unit cell
M = Molar mass of the element
a = Unit cell edge length
Nₐ = Avogadro's number (6.022 × 1023 atoms/mol)
Rearranging the formula, we get:
M = (Density × a3 × Nₐ) / Z
Step 2: Calculate the volume of the unit cell.
The volume of a cube can be calculated using the formula:
Volume = a3
Step 3: Calculate the number of unit cells in 0.5 kg of the element.
Given:
Mass of the element = 0.5 kg
Using the molar mass of the element, we can calculate the number of moles:
Number of moles = Mass / Molar mass
Step 4: Calculate the number of atoms in the unit cell.
The number of atoms in the unit cell for an FCC lattice is 4.
Step 5: Calculate the total number of atoms in 0.5 kg of the element.
Total number of atoms = Number of unit cells × Number of atoms per unit cell
Now let's calculate the values:
Step 1:
M = (8.0 g/cm3 × (300 × 10-12 m)3 × (6.022 × 1023 atoms/mol)) / Z
Step 2:
Volume = (300 × 10-12 m)3
Step 3:
Number of moles = 0.5 kg / Molar mass
Step 4:
Number of atoms per unit cell = 4
Step 5:
Total number of atoms = Number of unit cells × Number of atoms per unit cell
= (Number of moles × Avogadro's number) × 4
After calculating all the values and substituting them into the formulas, we find that the correct answer is option 'C': 92.59 × 1023 atoms.
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Community Answer
An element of density 8.0 g/cm3 forms an FCC lattice with unit cell ed...
Given,
Density (ρ) = 8.0 g/cm3
For FCC structure, Z = 4
Avogadro’s number (N0) = 6.02 x 1023
Edge length of the unit cell (a) = 300 x 10-10 cm
The density of the element (ρ) = (Z x M)/ (a3 x N0)
Therefore, the Molar Mass (M) = (ρ x a3 x N0)/(Z)
= (8.0 x 6.02 x 1023 x 27.0 x 10-24) /4
= 32.508 g.
Therefore, 32.508 g of the element contains 6.02 x 1023 atoms.
500 g of the element would contain = (6.02 x 1023 x 500)/ 32.508 = 92.59 x 1023 atoms.
Density (ρ) = 8.0 g/cm3
For FCC structure, Z = 4
Avogadro’s number (N0) = 6.02 x 1023
Edge length of the unit cell (a) = 300 x 10-10 cm
The density of the element (ρ) = (Z x M)/ (a3 x N0)
Therefore, the Molar Mass (M) = (ρ x a3 x N0)/(Z)
= (8.0 x 6.02 x 1023 x 27.0 x 10-24) /4
= 32.508 g.
Therefore, 32.508 g of the element contains 6.02 x 1023 atoms.
500 g of the element would contain = (6.02 x 1023 x 500)/ 32.508 = 92.59 x 1023 atoms.
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An element of density 8.0 g/cm3 forms an FCC lattice with unit cell edge of 300 pm. Calculate the number of atoms present in 0.5kg of the element.a)95 x 1023 atomsb)93.59 x 1023 atomsc)92.59 x 1023 atomsd)91.38 x 1023 atomsCorrect answer is option 'C'. Can you explain this answer?
Question Description
An element of density 8.0 g/cm3 forms an FCC lattice with unit cell edge of 300 pm. Calculate the number of atoms present in 0.5kg of the element.a)95 x 1023 atomsb)93.59 x 1023 atomsc)92.59 x 1023 atomsd)91.38 x 1023 atomsCorrect 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 An element of density 8.0 g/cm3 forms an FCC lattice with unit cell edge of 300 pm. Calculate the number of atoms present in 0.5kg of the element.a)95 x 1023 atomsb)93.59 x 1023 atomsc)92.59 x 1023 atomsd)91.38 x 1023 atomsCorrect 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 An element of density 8.0 g/cm3 forms an FCC lattice with unit cell edge of 300 pm. Calculate the number of atoms present in 0.5kg of the element.a)95 x 1023 atomsb)93.59 x 1023 atomsc)92.59 x 1023 atomsd)91.38 x 1023 atomsCorrect answer is option 'C'. Can you explain this answer?.
An element of density 8.0 g/cm3 forms an FCC lattice with unit cell edge of 300 pm. Calculate the number of atoms present in 0.5kg of the element.a)95 x 1023 atomsb)93.59 x 1023 atomsc)92.59 x 1023 atomsd)91.38 x 1023 atomsCorrect 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 An element of density 8.0 g/cm3 forms an FCC lattice with unit cell edge of 300 pm. Calculate the number of atoms present in 0.5kg of the element.a)95 x 1023 atomsb)93.59 x 1023 atomsc)92.59 x 1023 atomsd)91.38 x 1023 atomsCorrect 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 An element of density 8.0 g/cm3 forms an FCC lattice with unit cell edge of 300 pm. Calculate the number of atoms present in 0.5kg of the element.a)95 x 1023 atomsb)93.59 x 1023 atomsc)92.59 x 1023 atomsd)91.38 x 1023 atomsCorrect answer is option 'C'. Can you explain this answer?.
Solutions for An element of density 8.0 g/cm3 forms an FCC lattice with unit cell edge of 300 pm. Calculate the number of atoms present in 0.5kg of the element.a)95 x 1023 atomsb)93.59 x 1023 atomsc)92.59 x 1023 atomsd)91.38 x 1023 atomsCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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