Hello,
I have 10nm diameter gold nanoparticle colloids dispersed in the water having 5.7x1012 particles/ml. can you calculate the molarity of that solution and show the steps? the density of gold is 19.6 g/cm^3.
Binay Priyadarsan Nayak
The quantitative amount of a substance (mol) is a measure of how many elementary elements of a given substance (atom, molecule, electron ....) are contained in an object or sample. A nanoparticle is not an elementary element. It is impossible to calculate the mole of nanoparticles. So it is impossible to calculate the concentration of mol / l.
Yuri Mirgorod, Thanks for the suggestions. Then how to calculate mol/l of Nanoparticles?
Binay Priyadarsan Nayak
The hypothetical concentration of monosized gold nanoparticles can be calculated from the relation
1 mol/l ------ 6.02X10 to the 23rd degree of nanoparticles,
? mol/l ---------5.7X10 to the 15th degree of nanoparticles
. ?=0.00000000094 mol/l
Binay Priyadarsan Nayak A molarity is simply moles/L of the appropriate entity and is a concentration measurement. It usually refers to a solution in chemistry but doesn't need to. For example, see:
https://en.wikipedia.org/wiki/Molar_concentration
One example from this is:
The density of water is approximately 1000 g/L and its molar mass is 18.02 g/mol (or 1/18.02 = 0.055 mol/g). Therefore, the molar concentration of water is c(H2O) = 1000 g/L/18.02 g/mol ≈ 55.5 mol/L
OK, back to your question.
The density of gold is actually 19.3 g/cm3 not 19.6 g/cm3. Interestingly the atomic mass of Au is 196.97 so there's potential room for confusion...
In terms of your original question, calculate the mass of a single particle of gold based on the size (π/6).d3.ρ (where d is the diameter and ρ the density of the gold. If you use 19.3 g/cm3 for the gold density, then the diameter needs to be in cm). You know you have 5.7E12 particles/mL or 5.7E15 particles/L so you can then work out the mass of gold present in the system or in 1 L of the system. This is easily converted to molarity (mol.L-1) as a 1M Au system will contain 196.97 g/L of gold. Only issue is getting the units correct. If your density is in g/cm3 then your diameters need to be in cm (not nm).
So, 10 nm = 10E-9 m = 10E-7 cm
Mass of 1 particle = (0.523)*(10E-7)3*19.3 g
Mass of 5.7E15 particles (in 1 L) = 5.7E15*(0.523)*(10E-7)3*19.3 g
Concentration of Au is [5.7E15*(0.523)*(10E-7)3*19.3]/196.97 mol/L
Finally, in line with the comments above, you state 'can you calculate the molarity of that solution', then the system you describe is not a solution and therefore the answer to that question is 'No'. However, you can calculate the concentration of particles in mol/L as I have done above.
You may wish to redo the calculation with the diameter of the particles in smoots: There are 170.18 cm in 1 smoot. What is the diameter of your gold particles in smoots?
Binay Priyadarsan Nayak Also take a look at:
https://www.researchgate.net/post/How_to_calculate_the_concentration_of_gold_Nanoparticles
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