[fangs]

I want to prepare a solution of benzene in methanol, where methanol is the solvent. I want to prepare several solutions of different concentrations in the ppm range between 50ppm up to 600ppm. I am working with ppmw (ppm in weight basis).

I have worked out the numbers but I need to get it checked that my calculation procedure is correct before I can actually proceed in the lab.

For benzene the density is d=0.8765g/mL

1- If I take 1mL volume v of Benzene how much is that in terms of mass m? So by using the density I calculate the mass of benzene as: m=d*v=0.8765g/mL * 1mL = 0.8765g = 876.5mg (miligrams)

2- Now if I take this 1mL of Benzene and transfer it to 100mL of methanol (solvent) what is my resulting solution concentration in ppmw? ppmw= mg/L = 876.5mg Benzene * 1/100mL methanol * 1000mL/1L = 8765ppmw

3- So now by transferring 1mL of benzene into 100mL of methanol I have a STOCK benzene solution with concentration of 8765ppmw.

4- Now if I want to make a sample solution of lower concentration I will use the following equation CcVc=CdVd where C and V are concentration and Volume, respectively. While subscript c and d represent concentrated solution and dilution solution, respectively. (Cc is the concentration of my concentrated solution so in this case is my STOCK solution).

5- I want to make a 100mL benzene solution of 600ppmw (Cd=600ppmw, Vd=100mL) using my stock Cc=8765ppmw how much of my stock do I need Vc=?
Vc=Cd * Vd/Cc = 600 * 100/8765 = 6.84541mL

6- Finally I will take Vc=6.84541mL of my stock and add it to 100mL-6.84541mL=93.1546mL of methanol to get my solution of 100mL 600ppmw of benzene.

Is this correct?

[DrCMS]

I want to prepare a solution of benzene in methanol, where methanol is the solvent. I want to prepare several solutions of different concentrations in the ppm range between 50ppm up to 600ppm. I am working with ppmw (ppm in weight basis).

I have worked out the numbers but I need to get it checked that my calculation procedure is correct before I can actually proceed in the lab.

For benzene the density is d=0.8765g/mL

1- If I take 1mL volume v of Benzene how much is that in terms of mass m? So by using the density I calculate the mass of benzene as: m=d*v=0.8765g/mL * 1mL = 0.8765g = 876.5mg (miligrams)

2- Now if I take this 1mL of Benzene and transfer it to 100mL of methanol (solvent) what is my resulting solution concentration in ppmw? ppmw= mg/L = 876.5mg Benzene * 1/100mL methanol * 1000mL/1L = 8765ppmw

If you want ppm on a weight basis why are you using the volume of methanol rather than its weight?

[fangs]

Methanol is just my solvent, my main concern is the concentration of benzene to be in ppmw

[fangs]

Can anyone confirm my calculations please. I really would like a comment

[Borek]

If you want ppm w/w you need to use weight of benzene AND weight of methanol, what you have now is ppm w/v.

[fangs]

So how should I do this?

[Borek]

Use definition of ppmw - which means you have to calculate mass of methanol. That's what DrCMS suggested in the very first answer to your post.

Compare http://www.chembuddy.com/?left=concentration&right=ppm-ppb-ppt