Molality

Advantages

The primary advantage of using molality as a measure of concentration is that molality only depends on the masses of solute and solvent, which are unaffected by variations in temperature and pressure. In contrast, solutions prepared volumetrically (e.g. molar concentration or mass concentration) are likely to change as temperature and pressure change. In many applications, this is a significant advantage because the mass, or the amount, of a substance is often more important than its volume (e.g. in a limiting reagent problem).

Another advantage of molality is the fact that the molality of one solute in a solution is independent of the presence or absence of other solutes.

Problem areas

Unlike all the other compositional properties listed in "Relation" section (below), molality depends on the choice of the substance to be called “solvent” in an arbitrary mixture. If there is only one pure liquid substance in a mixture, the choice is clear, but not all solutions are this clear-cut: in an alcohol–water solution, either one could be called the solvent; in an alloy, or solid solution, there is no clear choice and all constituents may be treated alike. In such situations, mass or mole fraction is the preferred compositional specification.

Mass fraction

The conversions to and from the mass fraction, w₁, of the solute in a single-solute solution are

${\displaystyle w_{1}={\frac {1}{1+{\dfrac {1}{b_{1}M_{1}}}}},\quad b_{1}={\frac {w_{1}}{(1-w_{1})M_{1}}},}$

where b₁ is the molality and M₁ is the molar mass of the solute.

More generally, for an n-solute/one-solvent solution, letting b_i and w_i be, respectively, the molality and mass fraction of the i-th solute,

${\displaystyle w_{i}=w_{0}b_{i}M_{i},\quad b_{i}={\frac {w_{i}}{w_{0}M_{i}}}}$,

where M_i is the molar mass of the ith solute, and w₀ is the mass fraction of the solvent, which is expressible both as a function of the molalities as well as a function of the other mass fractions,

${\displaystyle w_{0}={\frac {1}{1+\displaystyle \sum _{j=1}^{n}{b_{j}M_{j}}}}=1-\sum _{j=1}^{n}{w_{j}}}$.

Substitution gives:

${\displaystyle w_{i}={\frac {b_{i}M_{i}}{1+\displaystyle \sum _{j=1}^{n}b_{j}M_{j}}},\quad b_{i}={\frac {w_{i}}{\left(1-\displaystyle \sum _{j=1}^{n}w_{j}\right)M_{i}}}}$.

Mole fraction

The conversions to and from the mole fraction, x₁ mole fraction of the solute in a single-solute solution are

${\displaystyle x_{1}={\frac {1}{1+{\dfrac {1}{M_{0}b_{1}}}}},\quad b_{1}={\frac {x_{1}}{M_{0}(1-x_{1})}}}$,

where M₀ is the molar mass of the solvent.

More generally, for an n-solute/one-solvent solution, letting x_i be the mole fraction of the ith solute,

${\displaystyle x_{i}=x_{0}M_{0}b_{i},\quad b_{i}={\frac {b_{0}x_{i}}{x_{0}}}={\frac {x_{i}}{M_{0}x_{0}}}}$,

where x₀ is the mole fraction of the solvent, expressible both as a function of the molalities as well as a function of the other mole fractions:

${\displaystyle x_{0}={\frac {1}{1+M_{0}\displaystyle \sum _{i=1}^{n}{b_{i}}}}=1-\sum _{i=1}^{n}{x_{i}}}$.

Substitution gives:

${\displaystyle x_{i}={\frac {M_{0}b_{i}}{1+M_{0}\displaystyle \sum _{j=1}^{n}b_{j}}},\quad b_{i}={\frac {x_{i}}{M_{0}\left(1-\displaystyle \sum _{j=1}^{n}x_{j}\right)}}}$.

Molar concentration (molarity)

The conversions to and from the molar concentration, c₁, for one-solute solutions are

${\displaystyle c_{1}={\frac {\rho b_{1}}{1+b_{1}M_{1}}},\quad b_{1}={\frac {c_{1}}{\rho -c_{1}M_{1}}}}$,

where ρ is the mass density of the solution, b₁ is the molality, and M₁ is the molar mass (in kg/mol) of the solute.

For solutions with n solutes, the conversions are

${\displaystyle c_{i}=c_{0}M_{0}b_{i},\quad b_{i}={\frac {b_{0}c_{i}}{c_{0}}}}$,

where the molar concentration of the solvent c₀ is expressible both as a function of the molalities as well as a function of the other molarities:

${\displaystyle c_{0}={\frac {\rho b_{0}}{1+\displaystyle \sum _{j=1}^{n}{b_{j}M_{j}}}}={\frac {\rho -\displaystyle \sum _{j=1}^{n}{c_{i}M_{i}}}{M_{0}}}}$.

Substitution gives:

${\displaystyle c_{i}={\frac {\rho b_{i}}{1+\displaystyle \sum _{j=1}^{n}{b_{j}M_{j}}}},\quad b_{i}={\frac {c_{i}}{\rho -\displaystyle \sum _{j=1}^{n}{c_{j}M_{j}}}}}$,

Mass concentration

The conversions to and from the mass concentration, ρ_solute, of a single-solute solution are

${\displaystyle \rho _{\mathrm {solute} }={\frac {\rho bM}{1+bM}},\quad b={\frac {\rho _{\mathrm {solute} }}{M\left(\rho -\rho _{\mathrm {solute} }\right)}}}$,

or

${\displaystyle \rho _{1}={\frac {\rho b_{1}M_{1}}{1+b_{1}M_{1}}},\quad b_{1}={\frac {\rho _{1}}{M_{1}\left(\rho -\rho _{1}\right)}}}$,

where ρ is the mass density of the solution, b₁ is the molality, and M₁ is the molar mass of the solute.

For the general n-solute solution, the mass concentration of the ith solute, ρ_i, is related to its molality, b_i, as follows:

${\displaystyle \rho _{i}=\rho _{0}b_{i}M_{i},\quad b_{i}={\frac {\rho _{i}}{\rho _{0}M_{i}}}}$,

where the mass concentration of the solvent, ρ₀, is expressible both as a function of the molalities as well as a function of the other mass concentrations:

${\displaystyle \rho _{0}={\frac {\rho }{1+\displaystyle \sum _{j=1}^{n}b_{j}M_{j}}}=\rho -\sum _{i=1}^{n}{\rho _{i}}}$.

Substitution gives:

${\displaystyle \rho _{i}={\frac {\rho b_{i}M_{i}}{1+\displaystyle \sum _{j=1}^{n}b_{j}M_{j}}}\quad b_{i}={\frac {\rho _{i}}{M_{i}\left(\rho -\displaystyle \sum _{j=1}^{n}\rho _{j}\right)}}}$.

Equal ratios

Alternatively, one may use just the last two equations given for the compositional property of the solvent in each of the preceding sections, together with the relationships given below, to derive the remainder of properties in that set:

${\displaystyle {\frac {b_{i}}{b_{j}}}={\frac {x_{i}}{x_{j}}}={\frac {c_{i}}{c_{j}}}={\frac {\rho _{i}M_{j}}{\rho _{j}M_{i}}}={\frac {w_{i}M_{j}}{w_{j}M_{i}}}}$,

where i and j are subscripts representing all the constituents, the n solutes plus the solvent.

Example of conversion

An acid mixture consists of 0.76, 0.04, and 0.20 mass fractions of 70% HNO₃, 49% HF, and H₂O, where the percentages refer to mass fractions of the bottled acids carrying a balance of H₂O. The first step is determining the mass fractions of the constituents:

${\displaystyle {\begin{aligned}w_{\mathrm {HNO_{3}} }&=0.70\times 0.76=0.532\\w_{\mathrm {HF} }&=0.49\times 0.04=0.0196\\w_{\mathrm {H_{2}O} }&=1-w_{\mathrm {HNO_{3}} }-w_{\mathrm {HF} }=0.448\\\end{aligned}}}$.

The approximate molar masses in kg/mol are

${\displaystyle M_{\mathrm {HNO_{3}} }=0.063\ \mathrm {kg/mol} ,\quad M_{\mathrm {HF} }=0.020\ \mathrm {kg/mol} ,\ M_{\mathrm {H_{2}O} }=0.018\ \mathrm {kg/mol} }$.

First derive the molality of the solvent, in mol/kg,

${\displaystyle b_{\mathrm {H_{2}O} }={\frac {1}{M_{\mathrm {H_{2}O} }}}={\frac {1}{0.018}}\ \mathrm {mol/kg} }$,

and use that to derive all the others by use of the equal ratios:

${\displaystyle {\frac {b_{\mathrm {HNO_{3}} }}{b_{\mathrm {H_{2}O} }}}={\frac {w_{\mathrm {HNO_{3}} }M_{\mathrm {H_{2}O} }}{w_{\mathrm {H_{2}O} }M_{\mathrm {HNO_{3}} }}}\quad \therefore b_{\mathrm {HNO_{3}} }=18.83\ \mathrm {mol/kg} }$.

Actually, b_H2O cancels out, because it is not needed. In this case, there is a more direct equation: we use it to derive the molality of HF:

${\displaystyle b_{\mathrm {HF} }={\frac {w_{\mathrm {HF} }}{w_{\mathrm {H_{2}O} }M_{\mathrm {HF} }}}=2.19\ \mathrm {mol/kg} }$.

The mole fractions may be derived from this result:

${\displaystyle x_{\mathrm {H_{2}O} }={\frac {1}{1+M_{\mathrm {H_{2}O} }\left(b_{\mathrm {HNO_{3}} }+b_{\mathrm {HF} }\right)}}=0.726}$,

${\displaystyle {\frac {x_{\mathrm {HNO_{3}} }}{x_{\mathrm {H_{2}O} }}}={\frac {b_{\mathrm {HNO_{3}} }}{b_{\mathrm {H_{2}O} }}}\quad \therefore x_{\mathrm {HNO_{3}} }=0.246}$,

${\displaystyle x_{\mathrm {HF} }=1-x_{\mathrm {HNO_{3}} }-x_{\mathrm {H_{2}O} }=0.029}$.

Osmolality

Osmolality is a variation of molality that takes into account only solutes that contribute to a solution's osmotic pressure. It is measured in osmoles of the solute per kilogram of water. This unit is frequently used in medical laboratory results in place of osmolarity, because it can be measured simply by depression of the freezing point of a solution, or cryoscopy (see also: osmostat and colligative properties).

Relation to apparent molar properties and activity coefficients

For concentrated ionic solutions the activity coefficient of the electrolyte is split into electric and statistical components.

The statistical part includes molality b, hydration index number h, the number of ions from the dissociation and the ratio r_a between the apparent molar volume of the electrolyte and the molar volume of water.

Concentrated solution statistical part of the activity coefficient is:^[6][7][8]

${\displaystyle \ln \gamma _{s}={\frac {h-\nu }{\nu }}\ln \left(1+{\frac {br_{a}}{55.5}}\right)-{\frac {h}{\nu }}\ln \left(1-{\frac {br_{a}}{55.5}}\right)+{\frac {br_{a}\left(r_{a}+h-\nu \right)}{55.5\left(1+{\frac {br_{a}}{55.5}}\right)}}}$.

From the molalities of the binary solutions

The molalities of the solutes in a ternary solution can be expressed also from molalities in the binary solutions and their masses:

${\displaystyle b_{1}={\frac {m_{11}}{M_{1}(m_{01}+m_{02})}}={\frac {n_{11}}{m_{01}+m_{02}}}}$,

${\displaystyle b_{2}={\frac {m_{22}}{M_{2}(m_{01}+m_{02})}}={\frac {n_{22}}{m_{01}+m_{02}}}}$.

The binary solution molalities are:

${\displaystyle b_{11}={\frac {m_{11}}{M_{1}m_{01}}}={\frac {n_{11}}{m_{01}}}}$,

${\displaystyle b_{22}={\frac {m_{22}}{M_{2}m_{02}}}={\frac {n_{22}}{m_{02}}}}$.

The masses of the solutes determined from the molalities of the solutes and the masses of water can be substituted in the expressions of the masses of solutions:

${\displaystyle m_{s1}=m_{01}+m_{11}=m_{01}(1+b_{11}M_{1})}$.

Similarly for the mass of the second solution:

${\displaystyle m_{s2}=m_{02}+m_{22}=m_{02}(1+b_{22}M_{2})}$.

One can obtain the masses of water present in the sum from the denominator of the molalities of the solutes in the ternary solutions as functions of binary molalities and masses of solution:

${\displaystyle m_{01}={\frac {m_{s1}}{1+b_{11}M_{1}}}}$,

${\displaystyle m_{02}={\frac {m_{s2}}{1+b_{22}M_{2}}}}$.

Thus the ternary molalities are:

${\displaystyle b_{1}={\frac {b_{11}m_{01}}{m_{01}+m_{02}}}={\frac {b_{11}}{1+{\frac {m_{02}}{m_{01}}}}}={\frac {b_{11}}{1+{\frac {m_{s2}}{m_{s1}}}{\frac {1+b_{11}M_{1}}{1+b_{22}M_{2}}}}}}$,

${\displaystyle b_{2}={\frac {b_{22}m_{02}}{m_{01}+m_{02}}}={\frac {b_{22}}{1+{\frac {m_{01}}{m_{02}}}}}={\frac {b_{22}}{1+{\frac {m_{s1}}{m_{s2}}}{\frac {1+b_{22}M_{2}}{1+b_{11}M_{1}}}}}}$.

For solutions with three or more solutes the denominator is a sum of the masses of solvent in the n binary solutions which are mixed:

${\displaystyle b_{1}={\frac {m_{11}}{M_{1}(m_{01}+m_{02}+m_{03}+...)}}={\frac {n_{11}}{m_{01}+m_{02}+..}}={\frac {b_{11}}{1+{\frac {m_{02}}{m_{01}}}+{\frac {m_{03}}{m_{01}}}+...}}}$,

${\displaystyle b_{2}={\frac {m_{22}}{M_{2}(m_{01}+m_{02}+m_{03}+...)}}={\frac {n_{22}}{m_{01}+m_{02}+...}}}$,

${\displaystyle b_{3}={\frac {m_{33}}{M_{3}(m_{01}+m_{02}+m_{03}+...)}}={\frac {n_{33}}{m_{01}+m_{02}+...}}}$.