Question

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

Answer 1

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

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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...