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### M1V1=M2V2

Posted: Thu Oct 03, 2019 12:13 am
I understand that the molarity equation is M1V1=M2V2, however the molarity equation is = to M=n/V. So, basically, I am asking about the idea that shouldnâ€™t the volume cancel out in the M1V1=M2V2. I know my thinking is wrong but I just need a plausible explanation for it.

### Re: M1V1=M2V2  [ENDORSED]

Posted: Thu Oct 03, 2019 12:18 am
This equation demonstrates that initial and final moles are always equal when diluting a solution. So, yes, when you do multiply it out, the units for volume cancel out, but you are left with the number of moles. This equation helps us calculate volume needed for dilutions or new concentrations using the idea that moles are conserved in a solution.

### Re: M1V1=M2V2

Posted: Thu Oct 03, 2019 12:21 am
M wouldn't cancel out here since M1V1 actually comes from manipulating the equation Molarity(M)=number of moles(n)/volume. Since we know that moles of solute remain the same when diluting a solution we manipulate M=n/v to be n=MV so that we could set up the equation M1V1=M2V2 to solve for unknowns in the dilution process. I'm not that good at explaining equation but hopefully this help!

### Re: M1V1=M2V2

Posted: Thu Oct 03, 2019 12:52 am
Yes! Technically the equation would cancel out and become n1=n2, which is what M1V1=M2V2 originally derived from. Sometimes the problem would only give you molarity and volume so you wouldn't use moles.