1. calculate the volume of 1 spherical particle

Vpart=4/3*pi*r^3 where r=0.0005cm

2. calculate the mass of 1 spherical particle taking into account the density of the particle material (5g/cm3)

Vpart*5g/1cm3=m-part

3. calculate number of particles in 10 mg= 0.01 g

nº-part= 0.01/m-part

4. calculate the concentration of particles in 5ml

nº-part/5ml

From this concentration you can easily calculate the number of particles in the desired volume.

I hope it helps!

Hyvää päivää. Kiitos kysymys.  As-Salaam-Alaikum... السَّلَامُ عَلَيْكُمْ‎‎

You can do this with an assumption of the shape of the particles (and, probably most important, getting the units correct and consistent)

The mass of a single particle is (pi/6)*d3*density, assuming spheres.  Working in the cgs system (with your sizes of 10 um and density of 5g/cm3):

Mass of a single particle is (pi/6)*(10 X 10-4)3*5 = X g ~ 2.6 X 10-9 g

You have 10 mg (= 0.01g) of material.  This represents 0.01/X numbers of particles = Y ~ 3.8 X 106

This number of particles are in 5 mL.  Thus the number of particles in 100 uL is Y*[100*10-6/5*10-3] = 0.02Y ~ 7.6 X 104

You need to check and confirm my math as it’s easy to mix units or make an exponent error.

Dear Dr. Lluís and Dr. Alan F Rawle!

I am so grateful for the valuable support.

Kiitos paljon. Thank you very much. :)

The other point that occurs to me is that particles of 10 microns in size with a density of 5 g/cm3 will be sitting at the bottom of your container or cuvette.  Carry out a Stokes' law calculation.  Thus you'll not have a homogeneous system to dilute as there will be segregation in the system.

Dear Dr. Alan F Rawle!

Yes the particles tend to start settling down in few tens of seconds.

Does the non-existence of a homogeneous system mean that dilution will be erroneous? In our case, we require just a good approximation of the no. of particles. The purpose of doing dilutions this way is to avoid inaccuracy and tedious job of weighting very small quantities (0.5-1 mg/sample) of the particles for a large no. of samples. We can do sonication just before taking out the volume fractions of the particles to ensure that the system is homogeneous.

Thank you for the comment.

Yes, if you take a specimen from a heterogeneous sample, then your estimate of the numbers (obviously not the size if they're monodisperse) of particles will be in even greater error.  It's easy to magnetically stir say 20 mL of suspension and withdraw 100 uL by pipet but harder to envision a situation where this can easily be accomplished with 5 mL of starting suspension.  Imagine the situation where all particles have settled and you withdraw a specimen from the top layer.  Again, carry out a Stokes' law calculation as shown in ASTM E2490-09(15).

Thank you very much Dr. Alan!

I am not sure of the logic or accuracy of your statement: "We can do sonication just before taking out the volume fractions of the particles to ensure that the system is homogeneous". 

We will test if it works this way. otherwise, we will switch back to weighing of the particles. These methods of calculating the particle no. will still be very useful for us. i.e. for the estimation of no. of particles in a 0.5 mg.

Thank you :)

very confusing right from beginning. For dry sample it is pure mathematics with assumptions. Otherwise you would have to do counting, which is of course tedious.

When dispersing and drawing a subsample there will be indeed a relatively large systematic error. This could be minimized by increasing viscosity, e.g. by addition of glycerol.