 # What is ebullioscopy?

THE ebullioscopy, one of the four colligative properties, studies the behavior of the boiling point of a solvent when receiving one solute non-volatile. The other colligative properties are tonoscopy, cryoscopy and osmoscopy.

Note: Non-volatile solute is any substance that has a high boiling point and low melting point and capable of dissolving in a certain solvent.

Generally speaking, when a non-volatile solute is added to a solvent, it makes it difficult for the solvent to evaporate. Thus, a higher temperature is needed to be able to evaporate the solvent. At ebullioscopy, this increase in the boiling point of the solvent is studied.

This difficulty caused by the solute in the evaporation of the solvent, that is, the rise in the boiling point of the solvent, is directly related to the type of solute present in the solution. Possible types of solute are:

• Ionic solute: when added to water, ionize or dissociates, populating the solution with ions. Examples: salt, base, acid.

• molecular solute:

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when added to water, it does not ionize, maintaining the molecular shape. Examples: glucose, sucrose.

The greater the number of particles in the solvent, the more intense the ebullioscopy, that is, the higher the boiling point of the solvent. Thus, in ionic solutions, the boiling point of water tends to be always higher than the boiling point of molecular solutions, as long as they are in the same concentration.

Formulas used in ebullioscopy calculations

To perform the calculations of the ebullioscopy, we have the following formulas:

• Formula for calculating boiling temperature variation

Δte = t-t2

In this formula, we calculate the variation in the boiling temperature by subtracting the boiling temperature of the solvent, existing in the solution, from the boiling temperature of the pure solvent.

Note: The acronym Δte can also be called solvent boiling point elevation.

• Formula for calculating the boiling temperature rise involving the molality

Δte = Ke. W

It is a formula that, to be used, depends on the knowledge of the ebullioscopy constant, which is related to the solvent present in the solution, and the molality (W). Each of these variables has a particular formula.

The Van't Hoff correction factor (i) may also appear in this formula, however, only if the non-volatile solute present is ionic.

Δte = Ke. W.i

Note: To determine the Van't Hoff correction factor, we need the degree of ionization or dissociation of the solute and the number of particles (q) ionized or dissociated by the solute when present in water.

• Formula for calculating the ebuliscopic constant (Ke)

Ke = RT2
1000.Lv

In this formula, we have the general gas constant (0.082), temperature (always worked in kelvin) and the latent heat of vaporization.

• Formula for calculating molality (W)

W = m1
M1.m2

In this formula, there is the use of the mass of the solute (m1 - always worked in grams), of the molar mass of the solute (M1) and the mass of the solvent (m2 – always worked in kilograms).

Note: Based on the knowledge of the molality formula, if we replace the W, present in Δte's formula, by its respective formula, we will have the following result:

Δte = Ke.m1
M1.m2

Example of application of formulas in the calculation of ebullioscopy

1st Example - (Uece) Following in the footsteps of French chemist François-Marie Raoult (1830-1901), researching the ebuliometric effect of solutions, a chemistry student dissolved 90 g of glucose (C6H12O6) in 400 g of water and heated the whole. Knowing that Ke in water = 0.52 ºC/mol, after some time, the initial boiling temperature found by him was: (Data: Molar mass of glucose = 180 g/mol)

a) 99.85 °C.

b) 100.15 °C.

c) 100.50 °C.

d) 100.65 °C.

Data provided by the exercise:

• m1= 90 g;

• m2 = 400 g or 0.4 kg (after dividing by 1000);

• Ke = 0.52;

• M1 = 180 g/mol;

• t =? (initial boiling temperature or boiling temperature of the solvent in the solution).

Note: The boiling temperature of water (t2) is 100 OÇ.

As the exercise provided the masses and the ebullioscopy constant, just use the data in the expression below:

t-t2 = Ke.m1
M1.m2

t-100 = 0,52.90
180.0,4

t-100 = 46,8
72

t-100 = 0.65

t = 0.65 + 100

t = 100.65 OÇ

2nd Example - (Uece) Calcium chloride (CaCl2) has wide industrial application in refrigeration systems, in cement production, in milk coagulation for cheese production, and is excellently used as a moisture controller. A calcium chloride solution used for industrial purposes has a molality 2 and a boiling point of 103.016 °C under a pressure of 1 atm. Knowing that the ebullioscopy constant of water is 0.52 °C, its apparent degree of ionic dissociation is:

a) 80%.

b) 85%.

c) 90%.

d) 95%.

Data provided by the exercise:

• Ke = 0.52;
• W = 2 moles;
• t = 103.016 (initial boiling temperature or boiling temperature of the solvent in the solution).

Note: The boiling temperature of water (t2) is 100 OÇ.

As the exercise provided data on ebullioscopy, such as Ke and molality, it is evident that we should use the following formula for ebullioscopy:

Δte = Ke. W

However, as the exercise asks for the degree of dissociation, we must work the above formula with the Van't Hoff correction factor (i):

Δte = Ke. W.i

Also, to calculate the degree, you will need to replace i with its expression, which is 1 + α.(q-1):

t-t2 = Ke. W.[1 + α.(q-1)]

103,016-100 = 0,52.2.[1+ α.(3-1)]

3,016 = 1,04.[1+ 2 α]

3,016 = 1,04 + 2,08α

3,016 – 1,04 = 2,08α

1,976 = 2,08α

1,976 = α
2,08

α = 0,95

Finally, just multiply the value found by 100 to determine the percentage:

α = 0,95.100

α = 95%

By Me. Diogo Lopes Dias

Source: Brazil School - https://brasilescola.uol.com.br/o-que-e/quimica/o-que-e-ebulioscopia.htm

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