# help with moles

by hamxa7
Tags: moles
 P: 4 can someone clear my concepts about these two equations ??? Equation 1... Concentration(g/dm3) = Concentration (Mol/Dm3) % Mass Of 1 Mole Of Solute Equation 2... Number Of Moles Of Solute=Concentration(Mol/Dm3) x Volume In Cm3 % 1000cm3 reply quickly.. got my chem school exam tommorow...
 Admin P: 22,364 Hard to tell what you want to clear not knowing what your concepts are. I guess by % in the second equation you mean division, although the same % in the first equation is placed where the multiplication should be: $$C [\frac {g} {dm^3}] = C [\frac {mol}{dm^3}] \times M_m [\frac {g} {mol}]$$ where Mm is a molar mass. So, what is your problem (apart from procrastination)?
 P: 4 I wanted to know how are they are evolved... Answering a problem regarding the equations woud be appreciated...
P: 22,364

## help with moles

You have to be more specific, at the moment I have no idea what you are interested in - you question is ambiguous and way too vague.

By evolved - do you mean derived? If so, the second is just a a molar concentration definition solved for n, and with a mL <-> L conversion factor added. First can be obtained combining definition of molar concentration and definition of molar mass.
 P: 227 1dm3=1000cm3. The 1000 is just a conversion factor. Equation triangles helped a lot with me. (But volume has to be in dm3) ......^ ..../mol\ /vol*conc\ You use an equation triangle by covering up the quantity you want to work out. That leaves an equation to work it out using the other two. Also: .....^ ../mass\ /mol*Mr\ and .....^ .../vol\ /mol*24\