In chemistry, when the molar masses are calculated and if they are higher or lower than the expected value are known as abnormal molar masses. These are calculated using the colligative properties. Colligative properties include; elevation of boiling point, lowering relatively of vapor pressure, freezing point depression, and ease of osmotic pressure. The name in itself has abnormal in it which suggests the abnormality of how the molar masses are being calculated, using the Van’t Hoff factor. Let’s find out.

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## Introduction toÂ Abnormal Molar Masses

To understand the abnormal molar masses, first, we need to understand the total number of particles left either after the association or dissociation of the solute molecules present in a solution or a solvent. To study and understand more about more about the abnormalities, Dutch chemist Van’t Hoff, discovered (i), which is known as the Van’t Hoff’s factor.

**Browse more Topics under Solutions**

- Colligative Properties and determination of Molar Mass
- Expressing Concentration of Solutions
- Ideal and Non-ideal Solutions
- Osmosis and Osmotic Pressure
- Solubility
- Types of Solutions
- Vapour Pressure of Liquid Solutions

This factor represents the extension of dissociation or association of a solute. This factor alone accounts for all the abnormalities in the molar mass. We will study about the Van’t Hoff’s factor later on, but first, we need to know about the association and dissociation of the solute particles.

## Association of the Solute Particles

In a solution, some solute molecules start to associate with each other. Thus, now there is less number of these particles inside a solution. The colligative properties will vary with a number of solute particles in the solution. As a result, they will now, start to decrease with the number of solute particles being less.

The molecular mass of the solute is inversely proportional to its colligative properties. Thus a decrease in colligative properties will lead to an increase in the molecular masses.

Example: acetic acid or ethanoic acid associate in the solution, when dissolved in benzene, dimerizes, because the number of particles and hydrogen bondings are reduced. This happens usually with the solvents which have low dielectric constants. This experiment increases the molar mass more than the expected value achieved theoretically.

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## Dissociation of Solute Particles

Contrary to the association, some solute particles will dissociate into more ions or particles when they are dissolved in the solution. Thus, this will increase the number of solute particles in the solution and thereby the colligative properties of the solution.

Now, the molar mass being inversely proportional to the colligative properties will reduce the molar mass in the solution. This will result in the lower number of molar mass than the expected value. Thus the molar mass will be abnormal molar mass.

Example: The weak electrolytes are dissolved in the water, ions are formed. Hydrofluoric acid will dissociate into fluorine anion and hydrogen cation.

## Van’t Hoff factor

To understand the concept of abnormalities, Van’t Hoff introduced a factor called Van’t Hoff factor. This factor is used to sort out the dissociation and association problem while calculating the molar mass of the solute. The factor is denoted by ‘i’ and is obtained when the normal mass or the expected value is divided by the abnormal mass.

Mathematically, the factor is represented by, i = Normal molar mass/Abnormal molar mass or i = observed or expected colligative properties/calculated colligative properties or i = Total moles in the particles after dissociation or association/total moles in the particles before dissociation or association

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## How to calculate i?

Firstly, we will write whether the solute is being dissociated or associated.

For **association**, Î±= n\( \frac{i-1}{(n – 1) } \)

where n is the no. of particles being associated andÂ Î± is the degree of association. The equation is derived as follows:

nÎ± â†’ Î±_{n}

Initial moles: 1 molÂ Â andÂ Â 0 mol

Moles at equilibrium: 1 –Â Î±Â Â andÂ \( \frac{Î±}{n } \)

Therefore, total moles = 1 – Î± +Â \( \frac{Î±}{n } \)

Thus,Â Î± = n\( \frac{1-i}{(n – 1) } \)

For solute which depicts the association, the Van’t Hoff factor will always be less than 1. For **dissociation**, i = 1 + (n – 1)Î±. Here, Î± (alpha) will be the degree of dissociation and n will be a number of dissociated particles. The above equation will be derived from,

Î±_{n} â†’Â nÎ±

Initial moles: 1 moleÂ andÂ Â 0 moles

At the equilibrium: 1-Î±Â andÂ Â nÎ±

Total no. at equilibrium: 1 –Â Î± + nÎ±

Therefore, i = 1 –Â Î± +Â Â \( \frac{nÎ±}{1 } \)

Thus,Â Î± = \( \frac{i-1}{n -1} \)

For solutes depicting dissociation, Van’t Hoff factor will always be greater than 1. For particles, that shows neither association or dissociation, the Van’t Hoff factor is considered as 1. Thus, after the inclusion of Van’t Hoff’s factor, the equation of the colligative properties will be,

**Boiling point elevation**, âˆ†T_{b}= i xÂ K_{b}x m**Vapour pressure of the relatively lowering solvent**,Â \( \frac{p_1Âº – p_1}{p_1ÂºÂ } \)Â = \( \frac{i.n_2}{n_1} \)**Freezing point depression**, âˆ†T_{f}= i x k_{f}Ã— m

## Solved Examples for You!

**Question 1:** The depression of freezing point of the hydrofluoric acid solution is -0.201CÂ° with molality being 0.10 m. Find the Van’t Hoff factor. Take off as 1.86CÂ°m-Â¹.

Solution: As discussed earlier, the solution of hydrofluoric acid will dissociate into fluorine anion and hydrogen cation. Given data,

âˆ†T_{f} = -0.201CÂ°

m = 0.10

K_{f} = 1.86 CÂ°m-Â¹

Taking into account Van’t Hoff factor, the formula for depression of freezing point is, âˆ†T_{f} = if Thus, 0.201 = i x 1.86 x 0.10 So, i = 1.08. Thus the Van’t Hoff factor value will be 1.08.

**Question 2:** Find the Van’t Hoff factor for Sr(OH)_{2.}

Solution: For Sr(OH)_{2}, when it dissolves, the solution is separated into two ions;Â OH^{âˆ’} ions and Sr^{2+}. It will break up into 2 OH ions and 1 Sr ion. Thus, it will break up into 3 ions and so it’s Van’t Hoff factor will be 3.Â ^{Â }

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