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Pt crystallizes in a face-centered cubic (fcc) lattice with all atoms at the lattice points. It has a density of 21.45 g/cm³ and an atomic weight of 195.08 amu. Calculate the length of a unit-cell edge. STRATEGY: We can calculate the mass of the unit cell from the atomic weight. Knowing the density and the mass of the unit cell, we can calculate the volume of a unit cell and then the edge length of a unit cell. 195.08 g Pt x 1 mol Pt 1 mol Pt 6.022 x 1023 Pt atoms = 3.239 x 10-22g Pt 1 Pt atom Since there are 4 atoms of Pt per fcc unit cell, the mass of the unit cell = 4 x 3.239 x 10-22 1.296 x 10-21 g Pt = unit cell The Volume of the unit cell is: V = m/d AND V = a³ V = (1.296 x 10-21 g Pt/cell)/(21.45 g/cm³) = 6.042 x 10-23 cm³/cell

Pt crystallizes in a face-centered cubic (fcc) lattice with all atoms at the lattice points. It has a density of 21.45 g/cm³ and an atomic weight of 195.08 amu. Calculate the length of a unit-cell edge. STRATEGY: We can calculate the mass of the unit cell from the atomic weight. Knowing the density and the mass of the unit cell, we can calculate the volume of a unit cell and then the edge length of a unit cell. 195.08 g Pt x 1 mol Pt 1 mol Pt 6.022 x 1023 Pt atoms = 3.239 x 10-22g Pt 1 Pt atom Since there are 4 atoms of Pt per fcc unit cell, the mass of the unit cell = 4 x 3.239 x 10-22 1.296 x 10-21 g Pt = unit cell The Volume of the unit cell is: V = m/d AND V = a³ V = (1.296 x 10-21 g Pt/cell)/(21.45 g/cm³) = 6.042 x 10-23 cm³/cell

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Pt crystallizes in a face-centered cubic (fcc) lattice with all atoms at the lattice points. It has a density of 21.45 g/cm³ and an atomic weight of 195.08 amu. Calculate the length of a unit-cell edge. STRATEGY: We can calculate the mass of the unit cell from the atomic weight. Knowing the density and the mass of the unit cell, we can calculate the volume of a unit cell and then the edge length of a unit cell. 195.08 g Pt x 1 mol Pt 1 mol Pt 6.022 x 1023 Pt atoms = 3.239 x 10-22g Pt 1 Pt atom Since there are 4 atoms of Pt per fcc unit cell, the mass of the unit cell = 4 x 3.239 x 10-22 1.296 x 10-21 g Pt = unit cell The Volume of the unit cell is: V = m/d AND V = a³ V = (1.296 x 10-21 g Pt/cell)/(21.45 g/cm³) = 6.042 x 10-23 cm³/cell
Transcribed Image Text:Pt crystallizes in a face-centered cubic (fcc) lattice with all atoms at the lattice points. It has a density of 21.45 g/cm³ and an atomic weight of 195.08 amu. Calculate the length of a unit-cell edge. STRATEGY: We can calculate the mass of the unit cell from the atomic weight. Knowing the density and the mass of the unit cell, we can calculate the volume of a unit cell and then the edge length of a unit cell. 195.08 g Pt x 1 mol Pt 1 mol Pt 6.022 x 1023 Pt atoms = 3.239 x 10-22g Pt 1 Pt atom Since there are 4 atoms of Pt per fcc unit cell, the mass of the unit cell = 4 x 3.239 x 10-22 1.296 x 10-21 g Pt = unit cell The Volume of the unit cell is: V = m/d AND V = a³ V = (1.296 x 10-21 g Pt/cell)/(21.45 g/cm³) = 6.042 x 10-23 cm³/cell
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