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**ABNORMAL MOLAR MASS**

Molecular masses of the solute can be easily determined with the help of colligative properties; relative lowering in vapour pressure, boiling point elevation, freezing point depression, and osmotic pressure easily. But, we derived the relation between the molecular mass of solute and colligative properties under some assumptions, which are:

- The solution is diluted that is, solvent is available in large amount in order to obey Raoult’s Law
- The solute neither went dissociation or association in the solution

So, how do we determine the molecular mass of solute in case the above assumptions are not valid? Basically, the abnormality is due to:

**(i) Association of Solute Particles**

- Some solute molecules start to associate inside the solution.
- This means now there are less number of solute particles in the solution.
- As colligative properties vary with solute particles in the solution, they will decrease along with the solute particles.
- As colligative properties are inversely proportional to the molecular mass of solute, we get a higher molar mass of the solute.
**For Example:**Ethanoic acid or acetic acid ( CH_{3}COOH) associate in solution to form a dimer due to hydrogen bonding.

**Image 2: Ethanoic Acid dimer due to hydrogen bonding**

**(ii) Dissociation of Solute Particles**

- Some solute molecules, generally electrolytes dissociate into two or more ions/particles when dissolved in a solution.
- This leads to increase in solute particles in the solution, thereby increase in colligative properties of solutions.
- As colligative properties and molecular mass of solute varies inversely, we get a lower molar mass of the solute.

NaCl dissociate into two ions when dissolved in water

Such a molar mass, whose value is higher or lower than its expected value is known as Abnormal Molar Mass. For Example, weak electrolytes dissociate in water to form ions, example Hydrofluoric Acid (HF), dissociate into hydrogen cation and fluorine anion.

**VAN’T HOFF FACTOR**

To determine colligative properties for solutions which undergo association and dissociation, a Dutch physical chemist Jacobus Henricus van’t Hoff in the year 1880 introduced the concept of Van’t Hoff Factor i to sort out association and dissociation problem while calculating the molar mass of solute.

The Van’t Hoff factor is denoted by ‘i’ and is obtained when we divide normal mass with an abnormal mass of solute. Mathematically, it is represented in three forms:

i = Normal Molar Mass/Abnormal Molar Mass

i = Observed Colligative property/Calculated Colligative property

i = Total Number of Moles of Particles after Association/Dissociation / Total Number of Moles of Particles before Association / Dissociation

Thus, the value of i depends upon the state of solute in the solution.

To determine colligative properties for solutions which undergo association and dissociation, a Dutch physical chemist Jacobus Henricus van’t Hoff in the year 1880 introduced the concept of Van’t Hoff Factor i to sort out association and dissociation problem while calculating the molar mass of solute.

The Van’t Hoff factor is denoted by ‘i’ and is obtained when we divide normal mass with an abnormal mass of solute. Mathematically, it is represented in three forms:

i = Normal Molar Mass/Abnormal Molar Mass

i = Observed Colligative property/Calculated Colligative property

i = Total Number of Moles of Particles after Association/Dissociation / Total Number of Moles of Particles before Association / Dissociation

Thus, the value of i depends upon the state of solute in the solution.

**Following cases become possible.**

- When,
**i = 1,**then the solute remains unaffected (i.e., normal) in solution. - When,
**i > 1,**then the solute undergoes dissociation in solution. - When,
**i < 1,**then the solute undergoes association in solution.

Thus, for

Potassium chloride (KCl) in aqueous solution, i is nearly equal to 2.

Barium chloride (BaCl_{2}) in aqueous solution, i is nearly equal to 3.

Benzoic acid (C_{6}H_{5}COOH) in benzene, i is nearly equal to 0.5.

Acetic acid (CH_{3}COOH) in benzene, i is nearly equal to 0.5.**CALCULATION OF I (VAN’T HOFF FACTOR)**

First, we write an equation of the solute being associated or dissociated**CASE 1: **WHEN SOLUTE UNDERGOES DISSOCIATION WHEN DISSOLVED IN SOLVENT**Degree of dissociation:** It is defined as the number of moles dissociated over the initial number of moles taken.**Van’t Hoff factor =** Total number of particles at equilibrium/Total initial number of particle

Let c moles per litre of an electrolyte A_{x}B_{y} be taken and let it dissociated as

A_{x}B_{y} →xA^{y+} + yB^{x–}

t = 0 C 0 0

c (1 − α) xcα ycα At equilibrium

i = concentration or number of moles after dissociation/concentration or number of moles before dissociation

i =[c - cα + xcα + ycα ]/c

or i = [1 - α + xα + yα ]/1

= 1 + [(x + y)–1]α

Let x + y = n = number of particles formed after dissociation

i = 1 + (n – 1)α

**Degree of Association may be defined as the number of molecules associated per unit volume over initial concentration**

i = Total number of particles or concentration after association/Total number of particles or concentration before association**Let n molecules of an electrolyte A undergo association**

then, nA → A_{n}

C 0 t = 0

C – C · α C.α/n At. equilibrium

**MODIFIED EQUATIONS FOR COLLIGATIVE PROPERTIES**

For solutes which undergo dissociation or association in solutions, the equations for the colligative properties are modified by inserting the van'tHoff's factor in them as follows:

**Example 1. Acetic acid (CH _{3}COOH) associates in benzene to form double molecules. 1.65 g of acetic acid when dissolved in 100 g of benzene raised the boiling point by 0.36˚C. Calculate the Van't Hoff factor and the degree of association of acetic acid in benzene. (K_{b} = 2.57 K kg **

Solution.

Total moles at equilibrium

∆T

α = 0.981

**Example 2. Derive a relationship between K _{b}, T_{b} and ∆H_{vap} for a liquid. **

Let us consider a pure liquid. At its boiling point, T

P˚ = P

Now let us consider a non-volatile solute dissolved in the liquid. When the solution reaches temperature T

Assuming the solution to be highly dilute, T

Since X

Ignoring n in comparison to N in the denominator

Multiplying by 1000 on both the sides, we get,

Similarly, K

Total number of moles per kg of solvent = 0.1003 (1 + 0.23)

ΔT_{f} = K_{f}m = 1.86 × 0.1003 × 1.23 = 0.23o C**Example. 4. At 300 K, the vapour pressure of an ideal solution containing one mole of A and 3 moles of B is 500 mm of Hg. At the same temperature, if one mole of B is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. Calculate the V.P. of A and B in their pure state.****Solution. **

When, one mole of B is further added to it,

Solving equations (i) and (ii), we get

P°_{A}= 440 mm, P°_{B}= 600 mm**Example 5. A decinormal solution of NaCl exerts an osmotic pressure of 4.6 atm. at 300 K. Calculate its degree of dissociation. (R = 0.082 L atm. K ^{-1}mol^{-1}).**

No. of moles of NaCl per litre of solution = 0.1

Osmotic pressure (p) = 4.6 atm., Temperature (T) = 300 K

Had NaCl not dissociated, then

p normal = CRT = 0.1 × 0.082 × 300 atm. = 2.46 atm.

But, pobs = 4.6 atm.

As per definition,

For the dissociation of an electrolyte producing n ions,

So, Percentage dissociation = 100 × a = 100 × 0.87 = 87%**Example 6. Vapour pressure of C _{6}H_{6} and C_{7}H_{8} mixture at 50˚C are given by: P = 179 X_{B} 92, where X_{B} is mole fraction of C_{6}H_{6}. Calculate (in mm): **

(A) Given, P = 179X

For pure, C

P°

For pure C

P°

(B) Now PM = P°

= 162.6 +36.8 = 199.4 mm

Moles of C_{6}H_{6} = 936/78 = 12

Moles of C_{7}H_{8} = 736/92 = 8

(C) Now mole fraction of C_{6}H_{6} in vapour phase of initial mixture (X'_{T})

Moles fraction of C_{7}H_{8} in vapour phase of initial mixture (X'_{L})

These vapours are taken out and condensed into liquid. The liquid is again brought to 50˚C to get again vapour-liquid equilibrium.

Thus, mole fraction of C_{6}H_{6} in vapour phase of initial mixture

= Mole fraction of C_{6}H_{6} in liquid phase on II mixture X'_{B}

Similarly, mole fraction of C_{7}H_{8} in vapour phase of initial mixture

= Mole fraction of C_{7}H_{8} in liquid phase on II mixture X'_{T}

Therefore,

= 271 × 0.815 + 92 × 0.185 mm

= 220.865 + 17.02 = 237.885 mm

New mole fraction of C_{6}H_{6} in vapour phase

New mole fraction of C_{7}H_{8} in vapour phase = 0.072.**Example 7. The freezing point depression of an HF (aq) solution with molality 0.10 m is -0.201˚C. What will be the Van’t Hoff factor? Take the value of Kf to be 1.86˚Cm ^{-1}**

HF (aq) → H

We have Δ T

The freezing point depression is Δ T

0.201 = i × 1.86 × 0.10

The value of Van’t Hoff factor will be 1.08

Given, concentration of solution = C = 1/1000 M = 10

Where α = degree of dissociation = 70% = 70/100 = 0.7

Total moles of after dissociation = (1 – α) + 3α + α = 1+ 3α

∴ i = 1 + 3α/1

i = 1+3α = 1+(3 × 0.7) = 3.1

Since osmotic pressure

π = iCRT

∴ π = 3.1 × 10^{–3} × 0.0821 × 300

= 0.31 × 3 × 0.0821 atm = 0.07635 atm**TRY YOURSELF!****Q.1. A 0.5 percent aqueous solution of Potassium Chloride was found to freeze at 0.24°C. Calculate the Van't Hoff factor and degree of dissociation of the solute at this concentration (K _{f} for water is 1.86).**

i =

= 1.923; i = 1+(n–1)α

1.923 = 1+ (2–1)α ⇒ 1.923 = 1+ α

α = 1.923 – 1 = 0.923

or α = 0.923 or 92.3%

Due to ionization of KCl

From question, degree of ionization = a = 0.86

∴ i = 1+ 0.86 = 1.86

For Osmotic pressure of glucose,

For 4% glucose solution,

Weight of glucose = 40 gm,

Volume of solution = V = 1L = 1dm^{3}

Molecular weight of glucose C_{6}H_{12}O_{6}= m = 180

where π_{glucose} = osmotic pressure of glucose

Similarly,

As both solutions are isotonic,

∴ π_{KCl} = π_{glucose}

= 8.9 gm

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