# What is the relationship between normality and molarity? Normality and molality? Mole fraction and molarity?

##### 1 Answer

Make sure to read this answer carefully... these are very conditional relationships.

You have asked for:

#"N"# #-># #"M"# #"N"# #-># #"m"# #chi -> "m"# (I added this one)#chi -> "M"#

**DISCLAIMER:** *LONG ANSWER!*

**NORMALITY TO MOLARITY**

The hardest one to consider is

Then, **normality** is defined with respect to the

We define *equivalence factor*, which is case-dependent...

#color(blue)(barul|stackrel(" ")(" ""N" = (c)/(f_(eq))" ")|)# ,

#c# in#"mol/L"#

Keep in mind that

**NORMALITY TO MOLALITY**

Well, this is not easy either. Normality is proportional to molarity

#c = "mols solute"/"L solution" = n_"solute"/V_"soln"#

#= color(green)((n_"solute")/(w_"soln") xx rho_"soln")# where

#rho# is the density in#"g/L"# and#w# is the mass in#"g"# .

#m = "mols solute"/"kg solvent" = n_"solute"/w_"solvent"#

#= "mols solute"/("kg solution - kg solute")#

#= color(green)(n_"solute"/(w_"soln" - w_"solute"))#

Since we already know how to get from normality to molarity... consider getting from molarity to molality:

#overbrace((n_"solute"rho_"soln")/(w_"soln"))^"Molarity" " vs. " overbrace(n_"solute"/(w_"soln" - w_"solute"))^"Molality"#

Only for ** sufficiently dilute solutions**, i.e. the mass of solvent overwhelms the mass of solute, do we have

#c ~~ m xx rho_"solvent"#

This is usually considered to be around ** sufficiently-dilute solutions**...

#color(blue)(barul|stackrel(" ")(" ""N" ~~ (m xx rho_"solvent")/(1000 "g"/"kg" xx f_(eq))" ")|)# ,

#m# in#"mol/kg"#

#rho# in#"g/L"#

**MOL FRACTION TO MOLALITY**

For a mol fraction, consider a simple *binary ideal mixture* (one solute in one solvent):

#chi_"solute" = n_"solute"/(n_"soln") = n_"solute"/(n_"solute" + n_"solvent")# ,where

#n_"solute"# is the mols of solute and#n_"soln" = n_"solute" + n_"solvent"# .

Refer back to the definition of molality (not to be confused with mass!!) to find:

#m = n_"solute"/w_"solvent"# ,

The mols of solute can be found from the mol fraction.

#chi_"solute"(n_"solute" + n_"solvent") = n_"solute"#

#chi_"solute"n_"solute" + chi_"solute"n_"solvent" = n_"solute"#

#chi_"solute"n_"solvent" = (1 - chi_"solute")n_"solute"#

#n_"solute" = (chi_"solute"n_"solvent")/(1 - chi_"solute")#

Substituting back into the expression for the molality:

#m = (chi_"solute")/(1 - chi_"solute") xx n_"solvent"/w_"solvent"#

But the mols divided by mass is one over the ** molar mass**,

#color(blue)(barul|stackrel(" ")(" "m = (1000 "g"/"kg")/M_"solvent" xx (chi_"solute")/(1 - chi_"solute")" ")|)# ,

#M# in#"g/mol"#

#0 < chi < 1#

If you need it the other way around, you can solve for

**MOL FRACTION TO MOLARITY**

And for molarity, it is analogous, by adding in the result from earlier... which, again, we defined for *sufficiently-dilute solutions!*

#c ~~ m xx rho_"solvent"#

Thus, for solutions approximately

#color(blue)(barul|stackrel(" ")(" "c ~~ (rho_"solvent")/M_"solvent" xx (chi_"solute")/(1 - chi_"solute")" ")|)# ,

#M# in#"g/mol"#

#rho# in#"g/L"#

#0 < chi < 1#

For this relationship, you should find that