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The edge length of unit cell of a metal having molecular weight 75g/mol is 5Å which crystallizes in cubic lattice. If the density is 2g/cm3, then, find the radius (in pm, rounded up to first decimal places) of metal atom:
Correct answer is between '216.0,217.0'. Can you explain this answer?
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The edge length of unit cell of a metal having molecular weight 75g/mo...
For bcc ; r = √3/2 a;
d = n * M/NAV * a3 or n = d *N AV *a3/M
=> n = 2 *6 *1023 (5 *10-8)3/75 = 2
Therefore Metal crystallizes in BCC structure and for a BCC lattice √3a = 4r
r = √3/4 a = √3 *5/4 = 2.165 Å = 216.5 pm
so the required answer is 217 pm.
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The edge length of unit cell of a metal having molecular weight 75g/mo...
To calculate the edge length of the unit cell, we need to first determine the molar mass of the metal in grams per mole.
Given:
Molecular weight of the metal = 75 g/mol
Now, we can use the formula for molar mass:
Molar mass = Mass / Number of moles
Rearranging the formula to solve for the mass:
Mass = Molar mass × Number of moles
Since we are given the molecular weight of the metal, we can assume that it represents the molar mass.
Therefore, the mass of the metal is 75 g.
Next, we can calculate the number of moles of the metal.
Number of moles = Mass / Molar mass
Substituting the given values:
Number of moles = 75 g / 75 g/mol = 1 mol
Since we are dealing with a unit cell, the number of atoms in the unit cell is given by Avogadro's number, which is approximately 6.022 × 10^23.
Therefore, the number of atoms in the unit cell is 6.022 × 10^23 atoms.
Now, we can calculate the volume of the unit cell using the formula:
Volume = (Edge length)^3
Given:
Edge length = 5 Å (angstroms)
Converting the edge length from angstroms to meters:
1 Å = 1 × 10^-10 m
Therefore, the edge length in meters is:
Edge length = 5 × 10^-10 m
Now, we can substitute the values into the formula:
Volume = (5 × 10^-10 m)^3 = 1.25 × 10^-28 m^3
Finally, we can calculate the density of the metal using the formula:
Density = Mass / Volume
Substituting the values:
Density = 75 g / (1.25 × 10^-28 m^3) = 6 × 10^28 g/m^3
Given:
Molecular weight of the metal = 75 g/mol
Now, we can use the formula for molar mass:
Molar mass = Mass / Number of moles
Rearranging the formula to solve for the mass:
Mass = Molar mass × Number of moles
Since we are given the molecular weight of the metal, we can assume that it represents the molar mass.
Therefore, the mass of the metal is 75 g.
Next, we can calculate the number of moles of the metal.
Number of moles = Mass / Molar mass
Substituting the given values:
Number of moles = 75 g / 75 g/mol = 1 mol
Since we are dealing with a unit cell, the number of atoms in the unit cell is given by Avogadro's number, which is approximately 6.022 × 10^23.
Therefore, the number of atoms in the unit cell is 6.022 × 10^23 atoms.
Now, we can calculate the volume of the unit cell using the formula:
Volume = (Edge length)^3
Given:
Edge length = 5 Å (angstroms)
Converting the edge length from angstroms to meters:
1 Å = 1 × 10^-10 m
Therefore, the edge length in meters is:
Edge length = 5 × 10^-10 m
Now, we can substitute the values into the formula:
Volume = (5 × 10^-10 m)^3 = 1.25 × 10^-28 m^3
Finally, we can calculate the density of the metal using the formula:
Density = Mass / Volume
Substituting the values:
Density = 75 g / (1.25 × 10^-28 m^3) = 6 × 10^28 g/m^3
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The edge length of unit cell of a metal having molecular weight 75g/mol is 5Å which crystallizes in cubic lattice. If the density is 2g/cm3, then, find the radius (in pm, rounded up to first decimal places) of metal atom:Correct answer is between '216.0,217.0'. Can you explain this answer?
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The edge length of unit cell of a metal having molecular weight 75g/mol is 5Å which crystallizes in cubic lattice. If the density is 2g/cm3, then, find the radius (in pm, rounded up to first decimal places) of metal atom:Correct answer is between '216.0,217.0'. 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 The edge length of unit cell of a metal having molecular weight 75g/mol is 5Å which crystallizes in cubic lattice. If the density is 2g/cm3, then, find the radius (in pm, rounded up to first decimal places) of metal atom:Correct answer is between '216.0,217.0'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The edge length of unit cell of a metal having molecular weight 75g/mol is 5Å which crystallizes in cubic lattice. If the density is 2g/cm3, then, find the radius (in pm, rounded up to first decimal places) of metal atom:Correct answer is between '216.0,217.0'. Can you explain this answer?.
The edge length of unit cell of a metal having molecular weight 75g/mol is 5Å which crystallizes in cubic lattice. If the density is 2g/cm3, then, find the radius (in pm, rounded up to first decimal places) of metal atom:Correct answer is between '216.0,217.0'. 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 The edge length of unit cell of a metal having molecular weight 75g/mol is 5Å which crystallizes in cubic lattice. If the density is 2g/cm3, then, find the radius (in pm, rounded up to first decimal places) of metal atom:Correct answer is between '216.0,217.0'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The edge length of unit cell of a metal having molecular weight 75g/mol is 5Å which crystallizes in cubic lattice. If the density is 2g/cm3, then, find the radius (in pm, rounded up to first decimal places) of metal atom:Correct answer is between '216.0,217.0'. Can you explain this answer?.
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