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

1 Answer
Aug 14, 2017

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 #"N" -> "M"#, in the sense that it is completely contextual, despite such a simplistic formula... For simplicity, we consider only acids and bases...

Then, normality is defined with respect to the #"H"^(+)# in a strong Bronsted or Arrhenius acid or the #"OH"^(-)# in a strong Arrhenius base. For instance, a #"1 N H"_2"SO"_4# solution is about #"1 M"# in #"H"^(+)#, and thus we have #"0.5 M H"_2"SO"_4#.

We define #f_(eq)#, the equivalence factor, which is case-dependent...

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

#c# in #"mol/L"#

Keep in mind that #c# is the molarity for #"mols solute/L solution"#. For #"H"_2"SO"_4#, #1/(f_(eq)) = 2#. So, #1/(f_(eq))# indicates the number of #"H"^(+)# dissociated per unit of the strong acid.

NORMALITY TO MOLALITY

Well, this is not easy either. Normality is proportional to molarity #c#, and there is no clear relationship with molality #m#...

#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 #w_"solvent" ~~ w_"soln"#. Only then can we say that #w_"solute"# is small enough so that #rho_"soln" ~~ rho_"solvent"# and we have:

#c ~~ m xx rho_"solvent"#

This is usually considered to be around #"0.01 molal"#, or #"0.01 mols solute/kg solvent"# or less. So, for 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, #M_"solvent"# in #"g/mol"# (not to be confused with molarity!!), so:

#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 #chi#...

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 #"0.01 mols/kg"# or lower:

#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 #0.98 < chi_"solute" < 1# or so...