1.6.2: Composition- Scale Conversions- Molality

For a solution in a single solvent, chemical substance 1, containing solute \(j\),

\[\text { Molaity, } \mathrm{m}_{\mathrm{j}}=\mathrm{n}_{\mathrm{j}} / \mathrm{w}_{1} \nonumber \]

Here \(\mathrm{n}_{j}\) is the amount of solute and \(\mathrm{w}_{1}\) is the mass of solvent [1]

Mole Faction and Molality

For the same system, the amount of solvent,

\[\mathrm{n}_{1}=\mathrm{n}_{\mathrm{j}} / \mathrm{m}_{\mathrm{j}} \, \mathrm{M}_{1} \nonumber \]

But mole fraction, \(\mathrm{x}_{\mathrm{j}}=\mathrm{n}_{\mathrm{j}} /\left(\mathrm{n}_{\mathrm{1}}+\mathrm{n}_{\mathrm{j}}\right)\)

\[\text { Thus } \quad x_{j}=\frac{m_{j} \, n_{1} \, M_{1}}{n_{1}+m_{j} \, n_{1} \, M_{1}} \quad \text { or } \quad x_{j}=\frac{m_{j} \, M_{1}}{1+m_{j} \, M_{1}} \nonumber \]

For dilute solutions, \(1>>m_{j} \, M_{1}\).

\[\text { Then } x_{j}=m_{j} \, M_{1} \nonumber \]

For water(\(\ell\)), \(\mathrm{M}_{1}=0.018 \mathrm{~kg} \mathrm{~mol}^{-1}\). From equation (c) \(x_{j}+x_{j} \, m_{j} \, M_{1}=m_{j} \, M_{1}\)

\[\text { Then } \quad x_{j}=m_{j} \, M_{1} \,\left(1-x_{j}\right) \quad \text { or } \quad m_{j}=x_{j} /\left[M_{1} \,\left(1-x_{j}\right)\right] \nonumber \]

For dilute solutions \(1-x_{j} \approx 1.0\). and we recover equation (d). In short, equations (c) and (e) provide exact conversions between \(\mathrm{m}_{j}\) and \(\mathrm{x}_{j}\) whereas equation (d) is only valid for dilute solutions.

Concentration and Molality

We consider a solution having volume \(\mathrm{V}(\mathrm{s} \ln )\). \(\text { Mass of solution }=\rho(s \ln ) \, V(s \ln )\) where (at defined \(\mathrm{T}\) and \(\mathrm{p}\)), density \(=\rho(\mathrm{sln})\). If amount of substance \(j\) in this solution is \(\mathrm{n}_{j}\) mol then mass of solute \(\mathrm{w}_{\mathrm{j}}=\mathrm{n}_{\mathrm{j}} \, \mathrm{M}_{\mathrm{j}}\), where \(\mathrm{M}_{j} =\) molar mass of solute.

Mass of solvent in system \(=\rho(s \ln ) \, V(s \ln )-n_{j} \, M_{j}\)

\[\text { Hence molality } \quad m_{j}=\frac{n_{j}}{\rho(s \ln ) \, V(s \ln )-n_{j} \, M_{j}} \nonumber \]

\[\text { and concentration } \quad c_{j}=\frac{n_{j}}{V(s \ln )} \nonumber \]

From (f) and (g) \(\mathrm{m}_{\mathrm{j}}=\frac{\mathrm{n}_{\mathrm{j}}}{\frac{\rho(\mathrm{s} \ln ) \, \mathrm{n}_{\mathrm{j}}}{\mathrm{c}_{\mathrm{j}}}-\mathrm{n}_{\mathrm{j}} \, \mathrm{M}_{\mathrm{j}}}=\frac{1}{\frac{\rho(\mathrm{s} \ln )}{\mathrm{c}_{\mathrm{j}}}-\mathrm{M}_{\mathrm{j}}}\)

\[\text { Or, } m_{j}=\frac{c_{j}}{\rho(s \ln )-M_{j} \, c_{j}} \nonumber \]

\[\text { Or, } \quad c_{j}=\frac{m_{j} \, \rho(s \ln )}{1+m_{j} \, M_{j}} \nonumber \]

For dilute solutions \(\rho(\mathrm{s} \ln )>>\mathrm{M}_{\mathrm{j}} \, \mathrm{c}_{\mathrm{j}}\).

\[\text { Then[2] } c_{j}(s \ln )=m_{j} \, \rho(s \ln ) \nonumber \]

Therefore the exact conversion is given by equations (h) and (i) which reduce to equation (j) for dilute solutions.

Molality and Concentration

An elegant conversion is possible between \(\mathrm{m}_{j}\) and \(\mathrm{c}_{j}\) scales. The volume of a simple solution is given by equation (k).

\[\mathrm{V}(\mathrm{aq})=\mathrm{n}_{1} \, \mathrm{V}_{1}^{*}(\ell)+\mathrm{n}_{\mathrm{j}} \, \phi\left(\mathrm{V}_{\mathrm{j}}\right) \nonumber \]

Here \(\phi\left(\mathrm{V}_{\mathrm{j}}\right)\) is the apparent molar volume of solute \(j\).

\[\text { Or, } \quad \mathrm{V}(\mathrm{aq}) / \mathrm{n}_{\mathrm{j}}=\left(\mathrm{n}_{1} / \mathrm{n}_{\mathrm{j}}\right) \,\left[\mathrm{M}_{1} / \rho_{1}^{*}(\ell)\right]+\phi\left(\mathrm{V}_{\mathrm{j}}\right) \nonumber \]

\[\text { Or, } \quad 1 / \mathrm{c}_{\mathrm{j}}=\left[1 / \mathrm{m}_{\mathrm{j}} \, \rho_{1}^{*}(\ell)\right]+\phi\left(\mathrm{V}_{\mathrm{j}}\right) \nonumber \]

\[\text { Then, } 1 / \mathrm{m}_{\mathrm{j}}=\left[\rho_{1}^{*}(\ell) / \mathrm{c}_{\mathrm{j}}\right]-\left[\phi\left(\mathrm{V}_{\mathrm{j}}\right) \, \rho_{1}^{*}(\ell)\right] \nonumber \]

Footnotes

[1] Amount of solvent, molar mass \(\mathrm{M}_{1}\), \(\mathrm{n}_{1}=\mathrm{w}_{1} / \mathrm{M}_{1}\) Then, \(\mathrm{m}_{\mathrm{j}}=\mathrm{n}_{\mathrm{j}} / \mathrm{n}_{1} \, \mathrm{M}_{1}\) Units \(\mathrm{m}_{\mathrm{j}}=\mathrm{n}_{\mathrm{j}} / \mathrm{n}_{1} \, \mathrm{M}_{1}=[\mathrm{mol}] /[\mathrm{mol}] \,\left[\mathrm{kg} \mathrm{mol}^{-1}\right]=\left[\mathrm{mol} \mathrm{kg}^{-1}\right]\)

[2] \(c_{j}=\left[\mathrm{mol} \mathrm{m}^{-3}\right]=\left[\mathrm{mol} \mathrm{kg}^{-1}\right] \,\left[\mathrm{kg} \mathrm{m}^{-3}\right]\)