The density of a mixture

- Two components of equal mass but different volume - 

Knowing the density of such a mixture is a common strange situation in engineering. This demonstration can be extended to more components provided the mass of each component is the same.

The situation

Consider that you have a solution, of a solid with certain density $\rho_S$ dilute in certain solvent of density $\rho_L$, in which both the solute and solvent are mixed in equal quantities. This is, the mass of the solute is equal to the mass of the solvent. What is the density of such a mixture?

An analytical approach

The solution to this case is not knew and can also be found elsewhere. However, in this post, some different flavor shall be given.

The density of the solute is,

$\rho_S=\dfrac{m_S}{V_S}$        Eq. (01)

while the density of the solvent is,

$\rho_L=\dfrac{m_L}{V_L}$        Eq. (02)

but since the mass of solute and solvent are equal, it follows,

$m=m_S=m_L$        Eq. (03)

Also, the density of the mixture should be,

$\rho=\dfrac{m_S+m_L}{V_S+V_L}=\dfrac{2m}{V_S+S_L}$       Eq. (04) 

The volumes, $V_S$ and $V_L$, in Eq .(04) can be determined from Eqs. (01-02) condering Eq. (03) as follows,

$V_S=\dfrac{m}{\rho_S}$        Eq. (05)

$V_L=\dfrac{m}{\rho_L}$        Eq. (06)

Substitution of Eqs. (05-06) into Eq. (04) produces,

$\rho=\dfrac{2m}{\dfrac{m}{\rho_S}+\dfrac{m}{\rho_L}}$

which can be simplified,

$\rho=\dfrac{2m}{\dfrac{\rho_L m+\rho_S m}{\rho_S \rho_L}}=\dfrac{2m}{\dfrac{m\left( \rho_L+\rho_S \right)}{\rho_S \rho_L}}=\dfrac{2\rho_S \rho_L}{\rho_L+\rho_S}$            Eq. (07)

Of course, Eq. (07) applies for the mixture of two liquids as well. Therefore, if you know the density of both substances, you can readily estimate the density of the mixture (provided the mass of the two components is the same).

A case to estimate the density of a mixture

Let us consider a slurry fluid made of the mixture of coal and water. The coal has specific gravity 2.5 while the slurry has composition 50% coal w/w. What is the density of the slurry?

Well, the slurry mixture is made of coal and water mixed in the same mass proportion. This is, the mass of the coal is the same as the mass of the water used to make the slurry. The volume is unknown: it can be the same but, who knows.

Starting with the specific gravity of the coal (solute), its density should be,

$\rho_S=\left(2.5\right) \left( 1000\,kg/m^3 \right)=2500\,kg/m^3$

For the water, let us consider that it is at room temperature (say, 25 °C). Similar temperatures around this one will give a very similar water density, so that you do not have to worry too much. Then,

$\rho_L=997.05\, kg/m^3$

Therefore, the density of such a slurry should be,

$\rho=1425.56\, kg/m3$

This is the end of the post. I hope you find it useful.

Ildebrando.

Composition variables for mixtures

 When composition of a mixture of several gases is to be considered it can be challenging to use all concepts to represent the right quantities. This is a brief explanation.

About mole $n$ and volume $V$

The number of moles for a pure substance is usually represented by $n$. However, for a mixture with several components the moles of each of these components are to be expressed as follows:

$n_1$, $n_2$, $n_3$,...

where the subscripts 1, 2, 3 indicate the component in the mixture. 

Important note: You should remember that moles are extensive variables which is not recommended for composition calculations purposes. You may go around this difficulty dividing $n$ by an intensive variable, which results in a new intensive variable.

On the other hand, you may also have volumetric concentrations [concentración volumétrica] $\bar{c}$ defined as:

$\bar{c}_i=\dfrac{n_i}{V}$

where $n_i$ stands for the mole of some component and $V$ for the volume of the mixture. When $\bar{c}$ is given in units such as mole/l or mole/dm$^3$ the volumetric concentration is also called molar concentration [molaridad].

Important note: volumetric concentration is recommended for liquid or solid mixtures since these change very little with temperature and pressure. However, the use of $\bar{c}_i$ is not advised for gas mixtures. 

Mole ratio $r_i$ and molal concentration $m_i$

This is another form for referring to composition in terms of moles of components in a mixture. Picking up the moles of component 1 as reference we may define the corresponding ratios $r_i$ for all others as:

$r_i=\dfrac{n_i}{n_1}$

On the hand, molal concentration $m_i$ is in fact a variation of the mass concentration (how it is expressed) of the single component  gas $m$. Remember that the mass $m$ can be defined as:

$m=nM$

where $M$ is the molar mass (molalidad) given in [mole/g]. However, the mass of a component in a gas mixture is defined as:

$m_i=\dfrac{n_i}{n_1M_1}=\dfrac{r_i}{M_1}$

In other words, the mass $m_i$ of a mixture component must be given in terms of the mass and moles of the other components.

Since mole and molality ratios are temperature and poressure independent, these are preferable for any physicochemical calcuation.

Mole fractions $x_i$

These are obtained dividing each of the number of moles ($n_1$, $n_2$,...), of each component, by the total number of moles $n_t$ (of the whole substance) which is defined as:

$n_t=n_1+n_2+n_3+...$

The mole fraction is then expressed as,

$x_i=\dfrac{n_i}{n_t}$

Also, the summation of the mole fractions is always equal to 1:

$x_1+x_2+x_3+...=1$

Important note: The composition of a mixture is determined when all mole fractions are given or can be determined. Since mole fractions are temperature and pressure independent, these are suitable, and possibly the most used, to describe the composition of any mixture.

Any question? Write in the comments and I shall try to help.

Other stuff of interest

• Some basic concepts on thermodynamics
• Who was Clausius?
• Units conversion
• LE 04 - Latch contact with encapsulated relay for turning on/off an AC bulb light
• About PID controllers
• Ways to control a process
• About pilot lights
• Solving the Colebrook equation

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Ildebrando.