Calculating the number of particles in a solution

O calculating the number of particles in a solution is a fundamental aspect for us to measure the colligative effect (osmoscopy, cryoscopy, ebullioscopy and tonoscopy) caused by the addition of a solute to a certain solvent.

The larger the amount of particles in the solute present in the solution, the more intense the colligative effect. The calculation of the number of particles mainly takes into account the nature of the solute that was added.

The classification of a solute in relation to its nature is carried out as follows:

• molecular solute

It is the solute incapable of suffering the phenomena of dissociation or ionization, regardless of the solvent to which it was added. Examples: glucose, sucrose, ethylene glycol etc.

Thus, as a molecular solute does not ionize or dissociate, if we add 15 molecules (particles) of it to the solvent, we will have 15 dissolved molecules.

• ionic solute

It is the solute that, when added to the solvent, undergoes the phenomenon of ionization (production of cations and anions) or dissociation (release of cations and anions). Examples: acids, bases, salts, etc.

So if we add 15 molecules of it to the solvent, we have 15 particles plus x particles.

Van't Hoff Correction Factor

Scientist Van't Hoff developed a formula to calculate the correction factor for number of particles of an ionic solute in a solution.

i = 1 + α.(q-1)

Being:

• i = Van't Hoff correction factor.

• α = degree of dissociation or ionization of the solute;

• q = number of particles obtained from dissociation or ionization of a solute;

The Van't Hoff correction factor must be used to multiply the value found for the number of particles in the solution. So, if, for example, the correction factor is 1.5 and the number of particles of the solute in the solution is 8.5.10²², we will have:

number of real particles of solute in solution = 1.5. 8,5.10²²

number of real particles of solute in solution = 12.75.10²²

or

number of real particles of solute in solution = 1,275.10²³

Examples of calculating the number of particles in a solution

Example 1: Calculation of the number of particles present in a solution containing 45 grams of sucrose (C₆H₁₂O₆) dissolved in 500 mL of water.

Exercise data:

• Solute mass = 45 grams;

• Solvent volume = 500 ml.

Do the following:

1^O Step: determine the molar mass of the solute.

To determine the mass of the solute, just multiply the atomic mass of the element by the number of atoms in it in the formula. Then add up all the results.

Carbon = 12.12 = 144 g/mol
Hydrogen = 1.22 = 22 g/mol
Oxygen = 16.11 = 196 g/mol

Molar mass =144 + 22 + 196
Molar mass = 342 g/mol

2^O Step: Calculate the number of particles using a rule of three involving the number of particles and the mass.

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To assemble the rule of three, we must remember that, in a molar mass, the mass is always related to Avogadro's constant, which is 6.02.10²³ entities (molecules or atoms, for example). Thus, as sucrose has molecules, as it is molecular (formed by a covalent bond), we have to:

342 grams of sucrose6.02.10²³ molecules
45 grams of sucrose x

342.x = 45. 6,02.10²³

x = 270,9.10²³
342

x = 0.79.10²³ molecules

or

x = 7.9.10²² molecules

Example 2: Calculate the number of particles present in a solution that contains 90 grams of potassium carbonate (K₂CO₃) dissolved in 800 ml of water. Knowing that the degree of dissociation of this salt is 60%.

Exercise data:

• Solute mass = 90 grams;

• Solvent volume = 800 ml;

• α = 60% or 0.6.

For determine the number of solute particles in that solution, it is interesting that the following steps are developed:

1^O Step: determine the molar mass of the solute.

To determine the mass of the solute, just multiply the atomic mass of the element by the number of atoms in it in the formula. Then add up all the results.

Potassium = 39.2 = 78 g/mol
Carbon = 12.1 = 12 g/mol
Oxygen = 16.3 = 48 g/mol

Molar mass =144 + 22 + 196
Molar mass = 138 g/mol

2^O Step: calculate the number of particles using a rule of three involving the number of particles and mass.

To assemble the rule of three, we must remember that, in a molar mass, the mass is always related to Avogadro's constant, which is 6.02.10²³ entities (ion-formula, molecules or atoms, for example). Thus, as the carbonate has an ion-formula because it is ionic (formed by an ionic bond), we have to:

138 grams of carbonate 6.02.10²³ molecules
90 grams of carbonate x

138.x = 90. 6,02.10²³

x = 541,8.10²³
138

x = 6.02.10²³ formula ions (particles)

3^O Step: calculate the number of particles (q) from the dissociation of the salt.

In potassium carbonate, we have the presence of two potassium atoms in the formula (K₂) and a unit of the anion CO₃. So the value of q for this salt is 3.

q = 3

4^O Step: calculate from the Van't Hoff correction factor.

i = 1 + α.(q-1)

i = 1 + 0.6.(3-1)

i = 1 + 0.6.(2)

i = 1 + 1.2

i = 2.2

5^O Step:determine the number of real particles present in the solution.

To determine the number of real particles in this solution, simply multiply the number of particles calculated in 2^O step by correction factor calculated in 4^O step:

y = 6.02.10²³. 2,2

y = 13,244.10²³ particles

By Me. Diogo Lopes Dias