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

Aug 14, 2017

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

• $\text{N}$ $\to$ $\text{M}$
• $\text{N}$ $\to$ $\text{m}$
• $\chi \to \text{m}$ (I added this one)
• $\chi \to \text{M}$

NORMALITY TO MOLARITY

The hardest one to consider is $\text{N" -> "M}$, 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 ${\text{H}}^{+}$ in a strong Bronsted or Arrhenius acid or the ${\text{OH}}^{-}$ in a strong Arrhenius base. For instance, a ${\text{1 N H"_2"SO}}_{4}$ solution is about $\text{1 M}$ in ${\text{H}}^{+}$, and thus we have ${\text{0.5 M H"_2"SO}}_{4}$.

We define ${f}_{e q}$, the equivalence factor, which is case-dependent...

$\textcolor{b l u e}{\overline{\underline{|}} \stackrel{\text{ ")(" ""N" = (c)/(f_(eq))" }}{|}}$,

$c$ in $\text{mol/L}$

Keep in mind that $c$ is the molarity for $\text{mols solute/L solution}$. For ${\text{H"_2"SO}}_{4}$, $\frac{1}{{f}_{e q}} = 2$. So, $\frac{1}{{f}_{e q}}$ indicates the number of ${\text{H}}^{+}$ 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 $m$...

$c = \text{mols solute"/"L solution" = n_"solute"/V_"soln}$

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

where $\rho$ is the density in $\text{g/L}$ and $w$ is the mass in $\text{g}$.

$m = \text{mols solute"/"kg solvent" = n_"solute"/w_"solvent}$

= "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 ${w}_{\text{solvent" ~~ w_"soln}}$. Only then can we say that ${w}_{\text{solute}}$ is small enough so that ${\rho}_{\text{soln" ~~ rho_"solvent}}$ and we have:

$c \approx m \times {\rho}_{\text{solvent}}$

This is usually considered to be around $\text{0.01 molal}$, or $\text{0.01 mols solute/kg solvent}$ or less. So, for sufficiently-dilute solutions...

$\textcolor{b l u e}{\overline{\underline{|}} \stackrel{\text{ ")(" ""N" ~~ (m xx rho_"solvent")/(1000 "g"/"kg" xx f_(eq))" }}{|}}$,

$m$ in $\text{mol/kg}$
$\rho$ in $\text{g/L}$

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 ${n}_{\text{solute}}$ is the mols of solute and ${n}_{\text{soln" = n_"solute" + n_"solvent}}$.

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

$m = {n}_{\text{solute"/w_"solvent}}$,

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

${\chi}_{\text{solute"(n_"solute" + n_"solvent") = n_"solute}}$

${\chi}_{\text{solute"n_"solute" + chi_"solute"n_"solvent" = n_"solute}}$

${\chi}_{\text{solute"n_"solvent" = (1 - chi_"solute")n_"solute}}$

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, ${M}_{\text{solvent}}$ in $\text{g/mol}$ (not to be confused with molarity!!), so:

$\textcolor{b l u e}{\overline{\underline{|}} \stackrel{\text{ ")(" "m = (1000 "g"/"kg")/M_"solvent" xx (chi_"solute")/(1 - chi_"solute")" }}{|}}$,

$M$ in $\text{g/mol}$
$0 < \chi < 1$

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!

$c \approx m \times {\rho}_{\text{solvent}}$

Thus, for solutions approximately $\text{0.01 mols/kg}$ or lower:

$\textcolor{b l u e}{\overline{\underline{|}} \stackrel{\text{ ")(" "c ~~ (rho_"solvent")/M_"solvent" xx (chi_"solute")/(1 - chi_"solute")" }}{|}}$,

$M$ in $\text{g/mol}$
$\rho$ in $\text{g/L}$
$0 < \chi < 1$

For this relationship, you should find that $0.98 < {\chi}_{\text{solute}} < 1$ or so...