# Help with moles

1. Oct 19, 2011

### hamxa7

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...

2. Oct 19, 2011

### Staff: Mentor

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)?

3. Oct 19, 2011

### hamxa7

I wanted to know how are they are evolved...
Answering a problem regarding the equations woud be appreciated...

4. Oct 19, 2011

### Staff: Mentor

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.

5. Oct 19, 2011

### jetwaterluffy

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\

Last edited: Oct 19, 2011
6. Oct 19, 2011

### Staff: Mentor

Bastardized way of dealing with equations when you don't know basic algebra. It is really much better to learn general method of solving the equation for an unknown, than to memorize particular approach to particular problem.