What is the Value of Avogadro's Number for Ru in hcp Packing?

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Discussion Overview

The discussion revolves around calculating Avogadro's number for the element Ruthenium (Ru) based on its hexagonal close-packed (hcp) structure, utilizing its density and unit cell volume. The context includes a homework problem involving theoretical and practical aspects of crystallography and atomic calculations.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a calculation for Avogadro's number based on the density of Ru and its unit cell volume, arriving at a value of 1.80e24.
  • Another participant questions the assumption that there are six atoms per unit cell in the hcp structure.
  • A third participant references common knowledge about atomic counts in different crystal structures, noting that for hcp, they were informed it is six atoms, but expresses uncertainty about this information.
  • A link to an external resource is provided, possibly for further clarification on the structure of solids.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the number of atoms per unit cell in hcp packing, indicating a disagreement on this foundational aspect of the problem.

Contextual Notes

There are unresolved assumptions regarding the correct number of atoms in the hcp unit cell, which affects the calculation of Avogadro's number. The discussion also highlights potential discrepancies between calculated and tabulated values.

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Homework Statement



The element Ru has hcp packing with a hexagonal unit cell. The density of Ru is 12410 kg/m^3 and the cell volume is 2.711 x 10-23 mL. Calculate the value of Avogadro's number to three significant figures based on these data. (Note: the value may differ from the tabular value).

Homework Equations



D= 12410 kg/m^3 , V= 2.711 x 10-23 mL

The Attempt at a Solution



6 atoms (m^3/12410 kg)(1/2.711e-23 mL)(1000 mL/1 L)(1 L/0.001 m^3)(1 kg/1000g)(101.07 g/1 mol)

I end up with 1.80e24 which is 3x the true value...this method works for every other structure but the hcp
 
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Are you sure it is six atoms per unit cell?
 
I was told that for the base its 2 for simple its 1 and for face its 4, but hcp no one told me except for a tutor and he said it was 6. Is it wrong?
 
http://www.chem.ox.ac.uk/icl/heyes/structure_of_solids/lecture1/Lec1.html
 
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