User:Jzana/Sodium Chloride-Water mixture

The Sodium-Chloride is well soluble in water. The data in this article is given in SI system (System of Quantities and Units), and uses tehe terminology issued by the IUPAC Green Book.

Volume of the solution
For example calculate the volume of a mixture of 20 % mass fraction:

The volume of the solution is less than the sum of the volume of the components, because of the Sodium-Chloride is an ionic solution. Really both of the Sodium and the Chloride ions connected to four water molecules by the Coulomb forces, causing a loss of volume.

The molar volume excess:
 * $$V^E = \frac{x_1 M_1+ x_2 M_2}{\rho}-(x_1 V_1 + x_2 V_2) $$

Where x1 and x2 the mol fractions of the components; M1 and M2 their molar masses; V1 and V2 their molar volume, ρ the density of the mixture

Calculation methods
Mass ratio from the mass fraction:
 * $$\zeta = \frac{w}{1-w}$$ (per solvent)

Mass fraction from the mass ratio:
 * $$w= \frac{1}{\zeta+1}$$ (per solution)

Amount of substance ratio from the amount fraction:
 * $$r = \frac{x}{1-x}$$ (per solvent)

amount fraction from the amount ratio:
 * $$x = \frac{r}{r+1}$$ (solution)

Amount fraction from the mass fraction:
 * $$x_1=\frac { \frac {w_1}{M_1} } { \frac {w_1}{ M_1} + \frac{1-w_1} {M_2}  } $$

Where x1 the amount fraction of the first component (the solute), w1 the mass fraction of the first component, M1 the molar mass of the first component, M2 the same for the second component (the solvent).

The γ mass concentration can be calculated from the w mass fraction using the density of the solution ρ: $$\gamma = w_1 \rho_{solution} $$, followed by calculating the amount of substance concentration by the molar mass of the solute: $$ c_1=\frac {w_1 \rho_{oldat} }{M_1} $$

The γ mass concentration is calculated from the ratio of the mass of the solute and the volume of the whole solution. In chemistry another calculation is performed by dividing only the volume of the solvent only: this is the mass per volume ratio. Their SI units are the same: kg/m3. In respect of the unknown volume excess it is hardly calculated.

Molality is the amount of substance of the solute divided by the mass of the solvent.

Knowing the m molality, we can calculate the r amount ratio by M, the molar mass of the solvent $$r=\frac{m}{M}$$. The next step to calculate the amount fraction (mole fraction).

Refractive index
References can be found: (Topac Inc.)
 * Brix: Data in sucrose calibrated refractometer

Viscosity
Viscosity of a sulution of 1,0661 mol/kg molality and amount fraction of 0,01884 mol/mol

More about the viscosity We can calculate the excess viscosity using the following equation:
 * $$\eta^E = \eta_{12} -(x_1 \eta_1 + x_2 \eta_2) $$

where x the amount fraction of components and η the dynamic viscosity

Surface tension


Referred in:. Measuring method: Another reference:

Specific heat capacity
Specific heat capacity and the viscosity: (graphics), lowered (when mass fraction growing) from 4300 J/kg K to 3300 J/kg K