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

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The discussion centers on calculating Avogadro's number for ruthenium (Ru) based on its density and hexagonal close-packed (hcp) unit cell volume. The density of Ru is given as 12410 kg/m^3, and the cell volume is 2.711 x 10^-23 mL. A calculation attempt yielded a result of 1.80e24, which is significantly higher than the accepted value, raising questions about the number of atoms per unit cell in hcp structures. There is uncertainty regarding whether hcp has six atoms per unit cell, as some participants reference different structures with varying atom counts. The thread highlights the importance of confirming the correct atomic count for accurate calculations in crystallography.
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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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