What do I mean by #"dimensionally"#?
Suppose we have a #1.00*g# mass of water. Now we know that #"Avogadro's number of water molecules"# has a mass of #18.01*g#, and we would typically write this as......
#"Molar mass of water"=18.01*g*mol^-1#
And if I wanted to work out the MOLAR quantity, I would perform the division by leaving the UNITS in......i.e.
#"Moles of water"=(1.00*g)/(18.01*g*mol^-1)#, and of course we can do some cancellation of units.....
#"Moles of water"=(1.00*cancelg)/(18.01*cancelg*mol^-1)#
#=0.0555*1/(mol^-1)=0.0555*1/(1/(mol))# because #x^-1-=1/x#, and
#=0.0555*1/(mol^-1)=0.0555*1/(1/(mol))=0.0555*mol#.
And as for volumes, we need a #"density"#, #rho#, the which for chemists is typically quoted as #g*mL^-1-=g*cm^-3#.
By definition, #rho="Mass"/"Volume"# i.e. mass per unit volume.
Here #rho_(H_2O)=("Molar quantity"xx"Molar mass")/(1*mL)#
#=(0.0555*cancel(mol)xx18.01*g*cancel(mol^-1))/(1*mL)#
#1*g*mL#; dimensionally consistent as required. Of course, the #"density"# needs to be measured......
If this does not help you will have to refine your question.
