There is an older discussion on this topic with detailed answers:

https://www.researchgate.net/post/How_do_you_calculate_the_concentration_of_gold_nanoparticles

Mehwish Fida

The only issue to deal with here is units and if the final concentration is to be expressed in μg/mL then it is sensible to have the density of Au in g/cm3 (19.3) and thus the size of the nanoparticle must be stated in cm. I assume by size you mean ‘diameter’ rather than radius. Please confirm as 'size' is an ambiguous term. In order to carry out the concentration we must assume some particle shape (or a shape factor), usually spherical, but no reason why cubes or other shapes (of calculable volume) can’t be used.

The mass of a single spherical particle is (π/6).d3.ρ where rho is the density of the particle. The shape factor here is π/6 (~ 0.524) but for a cube would be 1.0.

It’s easy to spreadsheet the calculation but I’ll show how to do the first one.

5 nm = 5*10-9 m = 5*10-7 cm

The are 1 million (or 106) μg in 1 g and the total mass of gold for a single particle will be multiplied by the number of particles to give the total gold concentration in the system.

Mass of a 5 nm single particle of gold = (π/6).d3.ρ = 0.524*(5*10-7)3*19.3*106 ~ 1.25E-12 μg

We have a stated range for the number of 5 nm particles ("4.92E+13 - 6.01E+13"), so we could take an arithmetic average, but we could simply use these data to generate a range of concentrations in line with the range in number. So, for the lower estimate of the particle number we’d have (1.25E-12)*(4.29E13) or ~ 54.2 μg/mL as the total mass of gold in the 5 nm system. Similarly for the higher number estimate (6.01E13) we can estimate a concentration of 75.9 ug/mL

I attach, in a screen dump of the Excel calculation, the other 2 examples:

You’ll see that the mass concentrations in μg/mL fall slightly as the size rises in line with the decreased numbers of particles. You should ask another person to repeat the same process to check my math or better still check the calculations yourself and construct your own spreadsheet(s).

Alan F Rawle, thank you so much for taking the time to answer my question much appreciated. Thank you Alireza and Kamyar.