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Help with molesby hamxa7
Tags: moles 
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#1
Oct1911, 03:05 AM

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... 


#2
Oct1911, 03:20 AM

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P: 23,599

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: [tex]C [\frac {g} {dm^3}] = C [\frac {mol}{dm^3}] \times M_m [\frac {g} {mol}][/tex] where M_{m} is a molar mass. So, what is your problem (apart from procrastination)? 


#3
Oct1911, 04:27 AM

P: 4

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


#4
Oct1911, 04:40 AM

Admin
P: 23,599

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. 


#5
Oct1911, 05:28 AM

P: 227

1dm^{3}=1000cm^{3}. The 1000 is just a conversion factor. Equation triangles helped a lot with me. (But volume has to be in dm^{3})
......^ ..../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\ 


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