Hi,

apparently your question is basic enough as to be answered without references. Pure water density is 1g/cm3, whereas your nanoparticles have a density different from 1. That is why weight fraction (solute mass/total mass) cannot be equivalent to volume fraction (solute volume/ total volume) because the equivalency between mass and volume in the solute is different from 1.

On the other hand, probably you can't precisely assess the solute volume since you don't know 'a priori' your particle shape and their size distribution. In a first aproach you can consider your nanoparticles as monodisperse spheres (V=4/3piR^3) to calculate their mean free path and the particle amount per cm3.

For instance, a 1% in weight colloid of Al2O3 nanoparticles in water (Al2O3 density = 3.95 g/cm3) means 1g of Al2O3 in 99 g of water. Therefore, you have 0.25 cm3 of Al2O3 in 99 cm3 of water, that is a 0.25 % in Volume of Al2O3. Furthermore, if your particles are spherical with a mean radius around R=50nm, then you have aproximately 4.7*10^14 particles in 0.25 cm^3. So that you have a colloid with 4.8*10^12 particles/cm^3.

Cheers!

Thank you Prof.Juan Marcos Sanz. The best an explanation that could be possible.

Dear Krishnan,

Basically,

-Mass fraction (wt.%) is [(mass of nanoparticle) / ( (mass of nanoparticle) + (mass of basefluid))] x 100.

-Volume fraction (vol.%), is [(mass of nanoparticle / density of nanoparticle) / ((mass of nanoparticle / density of nanoparticle) + (mass of basefluid / density of basefluid))] x 100

As you see volumetric fraction is a little bit more complex according to mass fraction. Generally, researchers use volumetric fraction in their research but some researches.

Best regards.