Molality

Amount of substance of solute divided by the mass of the solvent

Not to be confused with Molarity.

In chemistry, molality is a measure of the amount of solute in a solution relative to a given mass of solvent. This contrasts with the definition of molarity which is based on a given volume of solution.

A commonly used unit for molality is the moles per kilogram (mol/kg). A solution of concentration 1 mol/kg is also sometimes denoted as 1 molal. The unit mol/kg requires that molar mass be expressed in kg/mol, instead of the usual g/mol or kg/kmol.

## Definition

[edit]

The molality (b), of a solution is defined as the amount of substance (in moles) of solute, nsolute, divided by the mass (in kg) of the solvent, msolvent:[1]

b = n s o l u t e m s o l v e n t {\displaystyle b={\frac {n_{\mathrm {solute} }}{m_{\mathrm {solvent} }}}} .

In the case of solutions with more than one solvent, molality can be defined for the mixed solvent considered as a pure pseudo-solvent. Instead of mole solute per kilogram solvent as in the binary case, units are defined as mole solute per kilogram mixed solvent.[2]

## Origin

[edit]

The term molality is formed in analogy to molarity which is the molar concentration of a solution. The earliest known use of the intensive property molality and of its adjectival unit, the now-deprecated molal, appears to have been published by G. N. Lewis and M. Randall in the 1923 publication of Thermodynamics and the Free Energies of Chemical Substances.[3] Though the two terms are subject to being confused with one another, the molality and molarity of a dilute aqueous solution are nearly the same, as one kilogram of water (solvent) occupies the volume of 1 liter at room temperature and a small amount of solute has little effect on the volume.

## Unit

[edit]

The SI unit for molality is moles per kilogram of solvent.

A solution with a molality of 3 mol/kg is often described as "3 molal", "3 m" or "3 m". However, following the SI system of units, the National Institute of Standards and Technology, the United States authority on measurement, considers the term "molal" and the unit symbol "m" to be obsolete, and suggests mol/kg or a related unit of the SI.[4]

## Usage considerations

[edit]

### Advantages

[edit]

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

[edit]

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.

## Relation to other compositional quantities

[edit]

In what follows, the solvent may be given the same treatment as the other constituents of the solution, such that the molality of the solvent of an n-solute solution, say b0, is found to be nothing more than the reciprocal of its molar mass, M0 (expressed in the unit kg/mol):

b 0 = n 0 n 0 M 0 = 1 M 0 {\displaystyle b_{0}={\frac {n_{0}}{n_{0}M_{0}}}={\frac {1}{M_{0}}}} .

For the solutes the expression of molalities is similar:

b i = n i n 0 M 0 = x i x 0 M 0 = c i c 0 M 0 {\displaystyle b_{i}={\frac {n_{i}}{n_{0}M_{0}}}={\frac {x_{i}}{x_{0}M_{0}}}={\frac {c_{i}}{c_{0}M_{0}}}} .

The expressions linking molalities to mass fractions and mass concentrations contain the molar masses of the solutes Mi:

b i = n i n 0 M 0 = w i w 0 M i = ρ i ρ 0 M i {\displaystyle b_{i}={\frac {n_{i}}{n_{0}M_{0}}}={\frac {w_{i}}{w_{0}M_{i}}}={\frac {\rho _{i}}{\rho _{0}M_{i}}}} .

Similarly the equalities below are obtained from the definitions of the molalities and of the other compositional quantities.

The mole fraction of solvent can be obtained from the definition by dividing the numerator and denominator to the amount of solvent n0:

x 0 = n 0 n 0 + n 1 + n 2 + ∑ j = 3 n n j = 1 1 + n 1 n 0 + n 2 n 0 + ∑ j = 3 n n j n 0 {\displaystyle x_{0}={\frac {n_{0}}{n_{0}+n_{1}+n_{2}+\displaystyle \sum _{j=3}^{n}{n_{j}}}}={\frac {1}{1+{\frac {n_{1}}{n_{0}}}+{\frac {n_{2}}{n_{0}}}+\displaystyle \sum _{j=3}^{n}{\frac {n_{j}}{n_{0}}}}}} .

Then the sum of ratios of the other mole amounts to the amount of solvent is substituted with expressions from below containing molalities:

n i n 0 = b i M 0 {\displaystyle {\frac {n_{i}}{n_{0}}}=b_{i}M_{0}}

∑ i n n i n 0 = M 0 ∑ i n b i {\displaystyle \sum _{i}^{n}{\frac {n_{i}}{n_{0}}}=M_{0}\sum _{i}^{n}b_{i}}

giving the result

x 0 = 1 1 + M 0 b 1 + M 0 b 2 + M 0 ∑ i = 3 n b i = 1 1 + M 0 ∑ i = 1 n b i {\displaystyle x_{0}={\frac {1}{1+\displaystyle M_{0}b_{1}+M_{0}b_{2}+M_{0}\displaystyle \sum _{i=3}^{n}b_{i}}}={\frac {1}{1+M_{0}\displaystyle \sum _{i=1}^{n}b_{i}}}} .

### Mass fraction

[edit]

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

w 1 = 1 1 + 1 b 1 M 1 , b 1 = w 1 ( 1 − w 1 ) M 1 , {\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 bi and wi be, respectively, the molality and mass fraction of the i-th solute,

w i = w 0 b i M i , b i = w i w 0 M i {\displaystyle w_{i}=w_{0}b_{i}M_{i},\quad b_{i}={\frac {w_{i}}{w_{0}M_{i}}}} ,

where Mi 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,

w 0 = 1 1 + ∑ j = 1 n b j M j = 1 − ∑ j = 1 n w j {\displaystyle w_{0}={\frac {1}{1+\displaystyle \sum _{j=1}^{n}{b_{j}M_{j}}}}=1-\sum _{j=1}^{n}{w_{j}}} .

Substitution gives:

w i = b i M i 1 + ∑ j = 1 n b j M j , b i = w i ( 1 − ∑ j = 1 n w j ) M i {\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

[edit]

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

x 1 = 1 1 + 1 M 0 b 1 , b 1 = x 1 M 0 ( 1 − x 1 ) {\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 xi be the mole fraction of the ith solute,

x i = x 0 M 0 b i , b i = b 0 x i x 0 = x i M 0 x 0 {\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:

x 0 = 1 1 + M 0 ∑ i = 1 n b i = 1 − ∑ i = 1 n x i {\displaystyle x_{0}={\frac {1}{1+M_{0}\displaystyle \sum _{i=1}^{n}{b_{i}}}}=1-\sum _{i=1}^{n}{x_{i}}} .

Substitution gives:

x i = M 0 b i 1 + M 0 ∑ j = 1 n b j , b i = x i M 0 ( 1 − ∑ j = 1 n x j ) {\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)

[edit]

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

c 1 = ρ b 1 1 + b 1 M 1 , b 1 = c 1 ρ − c 1 M 1 {\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

c i = c 0 M 0 b i , b i = b 0 c i c 0 {\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:

c 0 = ρ b 0 1 + ∑ j = 1 n b j M j = ρ − ∑ j = 1 n c i M i M 0 {\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:

c i = ρ b i 1 + ∑ j = 1 n b j M j , b i = c i ρ − ∑ j = 1 n c j M j {\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

[edit]

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

ρ s o l u t e = ρ b M 1 + b M , b = ρ s o l u t e M ( ρ − ρ s o l u t e ) {\displaystyle \rho _{\mathrm {solute} }={\frac {\rho bM}{1+bM}},\quad b={\frac {\rho _{\mathrm {solute} }}{M\left(\rho -\rho _{\mathrm {solute} }\right)}}} ,

ρ 1 = ρ b 1 M 1 1 + b 1 M 1 , b 1 = ρ 1 M 1 ( ρ − ρ 1 ) {\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, ρi, is related to its molality, bi, as follows:

ρ i = ρ 0 b i M i , b i = ρ i ρ 0 M i {\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:

ρ 0 = ρ 1 + ∑ j = 1 n b j M j = ρ − ∑ i = 1 n ρ i {\displaystyle \rho _{0}={\frac {\rho }{1+\displaystyle \sum _{j=1}^{n}b_{j}M_{j}}}=\rho -\sum _{i=1}^{n}{\rho _{i}}} .

Substitution gives:

ρ i = ρ b i M i 1 + ∑ j = 1 n b j M j b i = ρ i M i ( ρ − ∑ j = 1 n ρ j ) {\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

[edit]

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:

b i b j = x i x j = c i c j = ρ i M j ρ j M i = w i M j w j M i {\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

[edit]

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:

w H N O 3 = 0.70 × 0.76 = 0.532 w H F = 0.49 × 0.04 = 0.0196 w H 2 O = 1 − w H N O 3 − w H F = 0.448 {\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

M H N O 3 = 0.063 k g / m o l , M H F = 0.020 k g / m o l , M H 2 O = 0.018 k g / m o l {\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,

b H 2 O = 1 M H 2 O = 1 0.018 m o l / k g {\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:

b H N O 3 b H 2 O = w H N O 3 M H 2 O w H 2 O M H N O 3 ∴ b H N O 3 = 18.83 m o l / k g {\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, bH2O 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:

b H F = w H F w H 2 O M H F = 2.19 m o l / k g {\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:

x H 2 O = 1 1 + M H 2 O ( b H N O 3 + b H F ) = 0.726 {\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} ,
x H N O 3 x H 2 O = b H N O 3 b H 2 O ∴ x H N O 3 = 0.246 {\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} ,
x H F = 1 − x H N O 3 − x H 2 O = 0.029 {\displaystyle x_{\mathrm {HF} }=1-x_{\mathrm {HNO_{3}} }-x_{\mathrm {H_{2}O} }=0.029} .

### Osmolality

[edit]

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

[edit]

Molality appears in the expression of the apparent (molar) volume of a solute as a function of the molality b of that solute (and density of the solution and solvent):

ϕ V ~ 1 = V − V 0 n 1 = ( m ρ − m 0 ρ 0 0 ) 1 n 1 = ( m 1 + m 0 ρ − m 0 ρ 0 0 ) 1 n 1 = ( m 0 ρ − m 0 ρ 0 0 ) 1 n 1 + m 1 ρ n 1 {\displaystyle {}^{\phi }{\tilde {V}}_{1}={\frac {V-V_{0}}{n_{1}}}=\left({\frac {m}{\rho }}-{\frac {m_{0}}{\rho _{0}^{0}}}\right){\frac {1}{n_{1}}}=\left({\frac {m_{1}+m_{0}}{\rho }}-{\frac {m_{0}}{\rho _{0}^{0}}}\right){\frac {1}{n_{1}}}=\left({\frac {m_{0}}{\rho }}-{\frac {m_{0}}{\rho _{0}^{0}}}\right){\frac {1}{n_{1}}}+{\frac {m_{1}}{\rho n_{1}}}} ,

ϕ V ~ 1 = 1 b 1 ( 1 ρ − 1 ρ 0 0 ) + M 1 ρ {\displaystyle {}^{\phi }{\tilde {V}}_{1}={\frac {1}{b_{1}}}\left({\frac {1}{\rho }}-{\frac {1}{\rho _{0}^{0}}}\right)+{\frac {M_{1}}{\rho }}} .

For multicomponent systems the relation is slightly modified by the sum of molalities of solutes. Also a total molality and a mean apparent molar volume can be defined for the solutes together and also a mean molar mass of the solutes as if they were a single solute. In this case the first equality from above is modified with the mean molar mass M of the pseudosolute instead of the molar mass of the single solute:

ϕ V ~ 12.. = 1 b T ( 1 ρ − 1 ρ 0 0 ) + M ρ {\displaystyle {}^{\phi }{\tilde {V}}_{12..}={\frac {1}{b_{T}}}\left({\frac {1}{\rho }}-{\frac {1}{\rho _{0}^{0}}}\right)+{\frac {M}{\rho }}} ,  M = ∑ y i M i {\displaystyle M=\sum y_{i}M_{i}}

y i = b i b T {\displaystyle y_{i}={\frac {b_{i}}{b_{T}}}} , yi,j being ratios involving molalities of solutes i,j and the total molality bT.

The sum of products molalities - apparent molar volumes of solutes in their binary solutions equals the product between the sum of molalities of solutes and apparent molar volume in ternary or multicomponent solution.[5]

ϕ V ~ 123.. ( b 1 + b 2 + b 3 + … ) = b 11 ϕ V ~ 1 + b 22 ϕ V ~ 2 + b 33 ϕ V ~ 3 + … {\displaystyle {}^{\phi }{\tilde {V}}_{123..}(b_{1}+b_{2}+b_{3}+\ldots )=b_{11}{}^{\phi }{\tilde {V}}_{1}+b_{22}{}^{\phi }{\tilde {V}}_{2}+b_{33}{}^{\phi }{\tilde {V}}_{3}+\ldots } .

### Relation to apparent molar properties and activity coefficients

[edit]

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 ra 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]

ln ⁡ γ s = h − ν ν ln ⁡ ( 1 + b r a 55.5 ) − h ν ln ⁡ ( 1 − b r a 55.5 ) + b r a ( r a + h − ν ) 55.5 ( 1 + b r a 55.5 ) {\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)}}} .

## Molalities of a ternary or multicomponent solution

[edit]

The molalities of solutes b1, b2 in a ternary solution obtained by mixing two binary aqueous solutions with different solutes (say a sugar and a salt or two different salts) are different than the initial molalities of the solutes bii in their binary solutions:

b 1 = m 11 M 1 ( m 01 + m 02 ) = n 11 m 01 + m 02 = b 11 1 + m 02 m 01 {\displaystyle b_{1}={\frac {m_{11}}{M_{1}(m_{01}+m_{02})}}={\frac {n_{11}}{m_{01}+m_{02}}}={\frac {b_{11}}{1+{\frac {m_{02}}{m_{01}}}}}} ,

b 2 = m 22 M 2 ( m 01 + m 02 ) = n 22 m 01 + m 02 = b 22 m 01 m 02 + 1 {\displaystyle b_{2}={\frac {m_{22}}{M_{2}(m_{01}+m_{02})}}={\frac {n_{22}}{m_{01}+m_{02}}}={\frac {b_{22}}{{\frac {m_{01}}{m_{02}}}+1}}} ,

b 11 = m 11 M 1 m 01 = n 11 m 01 {\displaystyle b_{11}={\frac {m_{11}}{M_{1}m_{01}}}={\frac {n_{11}}{m_{01}}}} ,

b 22 = m 22 M 2 m 02 = n 22 m 02 {\displaystyle b_{22}={\frac {m_{22}}{M_{2}m_{02}}}={\frac {n_{22}}{m_{02}}}} .

The content of solvent in mass fractions w01 and w02 from each solution of masses ms1 and ms2 to be mixed as a function of initial molalities is calculated. Then the amount (mol) of solute from each binary solution is divided by the sum of masses of water after mixing:

b 1 = 1 M 1 w 11 m s 1 w 01 m s 1 + w 02 m s 2 = 1 M 1 w 11 m s 1 ( 1 − w 11 ) m s 1 + ( 1 − w 22 ) m s 2 = 1 M 1 w 11 m s 1 m s 1 + m s 2 − w 11 m s 1 − w 22 m s 2 {\displaystyle b_{1}={\frac {1}{M_{1}}}{\frac {w_{11}m_{s1}}{w_{01}m_{s1}+w_{02}m_{s2}}}={\frac {1}{M_{1}}}{\frac {w_{11}m_{s1}}{(1-w_{11})m_{s1}+(1-w_{22})m_{s2}}}={\frac {1}{M_{1}}}{\frac {w_{11}m_{s1}}{m_{s1}+m_{s2}-w_{11}m_{s1}-w_{22}m_{s2}}}} ,

b 2 = 1 M 2 w 22 m s 2 w 01 m s 1 + w 02 m s 2 = 1 M 2 w 22 m s 2 ( 1 − w 11 ) m s 1 + ( 1 − w 22 ) m s 2 = 1 M 2 w 22 m s 2 m s 1 + m s 2 − w 11 m s 1 − w 22 m s 2 {\displaystyle b_{2}={\frac {1}{M_{2}}}{\frac {w_{22}m_{s2}}{w_{01}m_{s1}+w_{02}m_{s2}}}={\frac {1}{M_{2}}}{\frac {w_{22}m_{s2}}{(1-w_{11})m_{s1}+(1-w_{22})m_{s2}}}={\frac {1}{M_{2}}}{\frac {w_{22}m_{s2}}{m_{s1}+m_{s2}-w_{11}m_{s1}-w_{22}m_{s2}}}} .

Mass fractions of each solute in the initial solutions w11 and w22 are expressed as a function of the initial molalities b11, b22:

w 11 = b 11 M 1 b 11 M 1 + 1 {\displaystyle w_{11}={\frac {b_{11}M_{1}}{b_{11}M_{1}+1}}} ,

w 22 = b 22 M 2 b 22 M 2 + 1 {\displaystyle w_{22}={\frac {b_{22}M_{2}}{b_{22}M_{2}+1}}} .

These expressions of mass fractions are substituted in the final molalitaties:

b 1 = 1 M 1 1 1 w 11 + m s 2 w 11 m s 1 − 1 − w 22 m s 2 w 11 m s 1 {\displaystyle b_{1}={\frac {1}{M_{1}}}{\frac {1}{{\frac {1}{w_{11}}}+{\frac {m_{s2}}{w_{11}m_{s1}}}-1-{\frac {w_{22}m_{s2}}{w_{11}m_{s1}}}}}} ,

b 2 = 1 M 2 1 m s 1 w 22 m s 2 + 1 w 22 − w 11 m s 1 w 22 m s 2 − 1 {\displaystyle b_{2}={\frac {1}{M_{2}}}{\frac {1}{{\frac {m_{s1}}{w_{22}m_{s2}}}+{\frac {1}{w_{22}}}-{\frac {w_{11}m_{s1}}{w_{22}m_{s2}}}-1}}} .

The results for a ternary solution can be extended to a multicomponent solution (with more than two solutes).

### From the molalities of the binary solutions

[edit]

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

b 1 = m 11 M 1 ( m 01 + m 02 ) = n 11 m 01 + m 02 {\displaystyle b_{1}={\frac {m_{11}}{M_{1}(m_{01}+m_{02})}}={\frac {n_{11}}{m_{01}+m_{02}}}} ,

b 2 = m 22 M 2 ( m 01 + m 02 ) = n 22 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:

b 11 = m 11 M 1 m 01 = n 11 m 01 {\displaystyle b_{11}={\frac {m_{11}}{M_{1}m_{01}}}={\frac {n_{11}}{m_{01}}}} ,

b 22 = m 22 M 2 m 02 = n 22 m 02 {\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:

m s 1 = m 01 + m 11 = m 01 ( 1 + b 11 M 1 ) {\displaystyle m_{s1}=m_{01}+m_{11}=m_{01}(1+b_{11}M_{1})} .

Similarly for the mass of the second solution:

m s 2 = m 02 + m 22 = m 02 ( 1 + b 22 M 2 ) {\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:

m 01 = m s 1 1 + b 11 M 1 {\displaystyle m_{01}={\frac {m_{s1}}{1+b_{11}M_{1}}}} ,

m 02 = m s 2 1 + b 22 M 2 {\displaystyle m_{02}={\frac {m_{s2}}{1+b_{22}M_{2}}}} .

Thus the ternary molalities are:

b 1 = b 11 m 01 m 01 + m 02 = b 11 1 + m 02 m 01 = b 11 1 + m s 2 m s 1 1 + b 11 M 1 1 + b 22 M 2 {\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}}}}}} ,

b 2 = b 22 m 02 m 01 + m 02 = b 22 1 + m 01 m 02 = b 22 1 + m s 1 m s 2 1 + b 22 M 2 1 + b 11 M 1 {\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:

b 1 = m 11 M 1 ( m 01 + m 02 + m 03 + . . . ) = n 11 m 01 + m 02 + . . = b 11 1 + m 02 m 01 + m 03 m 01 + . . . {\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}}}+...}}} ,

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

b 3 = m 33 M 3 ( m 01 + m 02 + m 03 + . . . ) = n 33 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}+...}}} .

## See also

[edit]

* Molarity

## References

[edit]

1. ^ IUPAC, Compendium of Chemical Terminology, 5th ed. (the "Gold Book") (2025). Online version: (2006-) "molality". doi:10.1351/goldbook.M03970
2. ^ Sangster, James; Teng, Tjoon-Tow; Lenzi, Fabio (1976). "Molal volumes of sucrose in aqueous solutions of NaCl, KCl, or urea at 25°C". Journal of Solution Chemistry. 5 (8): 575–585\. doi:10.1007/BF00647379. S2CID 95559765.
3. ^ www.OED.com. Oxford University Press. 2011.
4. ^ "NIST Guide to SI Units". sec. 8.6.8. Retrieved 2007-12-17.
5. ^ Harned Owen, Physical Chemistry of Electrolytic Solutions, third edition 1958, p. 398-399
6. ^ Glueckauf, E. (1955). "The influence of ionic hydration on activity coefficients in concentrated electrolyte solutions". Transactions of the Faraday Society. 51: 1235. doi:10.1039/TF9555101235.
7. ^ Glueckauf, E. (1957). "The influence of ionic hydration on activity coefficients in concentrated electrolyte solutions". Transactions of the Faraday Society. 53: 305. doi:10.1039/TF9575300305.
8. ^ Kortüm, G. (1960). "The Structure of Electrolytic Solutions, herausgeg. von W. J. Hamer. John Wiley & Sons, Inc., New York; Chapman & Hall, Ltd., London 1959. 1. Aufl., XII, 441 S., geb. $ 18.50". Angewandte Chemie. 72 (24): 97. Bibcode:1960AngCh..72.1006K. doi:10.1002/ange.19600722427. ISSN 0044-8249.

Look up molality in Wiktionary, the free dictionary.

* v
* t
* e

Mole concepts

Constants|

* Avogadro constant
* Boltzmann constant
* Gas constant
* Faraday constant
* Molar mass constant

Physical quantities|

* Mass concentration
* Molar concentration
* Molality
* Mass
* Volume
* Density
* Mole fraction
* Mass fraction
* Amount of substance
* Molar mass
* Atomic mass
* Particle number
* Pressure
* Thermodynamic temperature
* Molar volume
* Specific volume

Laws|

* Charles's law
* Boyle's law
* Gay-Lussac's law
* Combined gas law
* Avogadro's law
* Ideal gas law

* v
* t
* e

Chemical solutions

Solution|

* Ideal solution
* Aqueous solution
* Solid solution
* Buffer solution
* Flory–Huggins
* Mixture
* Suspension
* Colloid
* Phase diagram
* Phase separation
* Eutectic point
* Alloy
* Saturation
* Supersaturation
* Serial dilution
* Dilution (equation)
* Apparent molar property
* Miscibility gap

Concentration
and related quantities|

* Molar concentration
* Mass concentration
* Number concentration
* Volume concentration
* Normality
* Molality
* Mole fraction
* Mass fraction
* Isotopic abundance
* Mixing ratio
* Ternary plot
* Total dissolved solids

Solubility|

* Solubility equilibrium
* Solvation
* Solvation shell
* Enthalpy of solution
* Lattice energy
* Raoult's law
* Henry's law
* Solubility table (data)
* Solubility chart
* Miscibility

Solvent|

* (Category)
* Acid dissociation constant
* Protic solvent
* Polar aprotic solvent
* Inorganic nonaqueous solvent
* Solvation
* List of boiling and freezing information of solvents
* Partition coefficient
* Polarity
* Hydrophobe
* Hydrophile
* Lipophilic
* Amphiphile
* Lyonium ion
* Lyate ion

*[v]: View this template
*[t]: Discuss this template
*[e]: Edit this template