Not sure what you mean by the bulk concentration. The molar mass of Na2Ti6O13 is 541.174 g/mol.  The atomic weight of 2 X Na is 45.98.  Hence the molar concentration of Na is 45.98/541.174 `~ 8.5 mol%

Physicists are always baffled by moles. Assuming that the lattice is of low symmetry and the unit cell is in fact primitive cell this is just 2/V_{unit}. 

Obviously volume and mass are linked via density and density is calculable from XRD . See attached. Given your measurements I make the theoretical density 2.11 g/cm3.  Please can someone confirm my calculation?

According to the link the z-number decribes the number of chemical formula per unit cell. So you have 4 Na atoms per unit cell; this amount is in units of mol:

4Na atoms = 4/A = 6,641* 10-24mol , with A= 6,023*1023atoms/mol, as the Avogadro constant 

Your volume unit of 106pm3 is equal to 10-24cm3. (it is equal to a cubic Angstrom)

So your concentration C(Na) is C = 6,641*10-24mol/( 512,18*10-24cm3)=0,013mol/cm3

http://www.uni-saarland.de/fileadmin/user_upload/Professoren/fr84_ProfMuecklich/lehre/crystallography/Lecture1-lattice-Indice.pdf

Dear Alan,

sorry I didn't recognized your second answer, but here is my reply:

 a) the formular for calculating the density is ok. The 'number of molecules per unit cell' is the z-number mentioned above.

According to your data with respect to molecular/ atomic masses given above and the cell volume given in the primary question I end up in contradiction to your number (2,11g/cm³) with:

b) mass density of Na is 0,298g/cm³   and

c) total mass density 3,51g/cm³.