Abnormal Molecular mass: Vant’s Hoff Factor

What is called Abnormal Molecular Mass?

The molecular mass of a solute present in a solution calculated with the help of colligative properties is different from its calculate molecular mass is called Abnormal molecular mass or Observed Molecular Mass.

• It is denoted by M_o.
• For an ideal solution, M_o = M_c (Where M_c is calculated molecular mass)
• In case of dissociating molecules, M_o < M_c ( Ex. Solution of NaCl in H₂O)
• In case of associating molecules, M_o > M_c (Ex. Solution of CH₃COOH in water)
• This phenomenon takes place when solution is non-ideal that means solution is not dilute or solutes undergo dissociation or association.

Vant’s Hoff Factor in Abnormal Molecular Mass

What is Van’s Hoff Factor ?

The ratio of the experimental value of the colligative properties to the calculated value of the colligative properties is called Van’t Hoff Factor. It is denoted by ‘i’.

The formula of Van’t Hoff Factor can be written as: General formula:$$i = \frac{\text{Observed colligative property}}{\text{Theoretical colligative property}}$$

For dissociation:$$i = 1 + \alpha (n – 1)$$

As we know that, Colligative Properties is inversely proportional to molecular mass hence $$i =\frac{Mc}{Mo}$$

Calculation of Van’t Hoff Factor in case of Dissociation:$$i =\frac{\text{No. of moles after dissociation}}{\text{No, of moles taken}}$$In case of strong electrolytic solute:
i = No. of moles after dissociation of solute and as α = 1
For example, In case of NaCl, i = 2 As NaCl → Na⁺ + Cl^–
In case of K₂SO₄, i = 3 As K₂SO₄ → 2K⁺ + SO4^2- .

In case of weak electrolytes: Suppose a molecule of a weak electrolyte is allowed to dissolve in a suitable solvent, it produces ‘n’ ions and its degree of dissociation be ‘α’. Then Vant’s Hoff factor can be calculated in the following way:

A → n₁B + n₂C + n₃D + —— or n₁ + n₂ + n₃ + —- = n

Since $$i =\frac{\text{No. of moles after dissociation}}{\text{No, of moles taken}}$$

And $$i =\frac{1 – α + n α}{1}$$ $$=\frac{1 – α (n – 1)}{1}$$ As we know that $$i =\frac{Mc}{Mo}$$ Hence, $$ α =\frac{Mc – Mo}{Mo (n – 1)}$$

Some important numericals regarding Van’s Hoff equation:

Question No. 1. The freezing point depression of 0.1 molal solution of acetic acid in benzene is 0.256 K. Kf for benzene is 5.12 K kg mol ^-1. What conclusion can you draw about the molecular of acetic acid in benzene ?

Solution: Here, we are given that m = 0.1mol kg^-1and K_f = 5.12 K kg mol^-1.
∴ Theoretical calculated value of ΔT_f  will be ΔT_f  = K_f × m = 5.12 K kg mol^-1 × 0.1mol kg^-1
= 0.512 K. As observed (experimental) value of ΔT_f = 0.256 K (Given)

Thus, the observed value is half of the theoratical value. This data suggests that acetic acid id doubly associated in benzene. Since the observed value depends upon the number of particles actually present in the solution.

Question No. 2. 0.5 g KCl was dissolved in 100 g water and the solution originally at 20^oC, froze at – 0.24^oC. Calculate the percentage ionization of salt. K_f per 100 g of water = 1.86^oC.

Solution:

Given Data:
\begin{aligned}
\text{Mass of KCl} &= 0.5 \text{ g} \\
\text{Mass of water} &= 100 \text{ g} \\
\text{Initial temperature} &= 25^\circ C \\
\text{Freezing point of solution} &= -0.24^\circ C \\
K_f &= 1.86^\circ C \cdot \text{kg}^{-1} \\
\text{Molar mass of KCl} &= 74.5 \text{ g/mol}
\end{aligned}

Step 1: Calculate Freezing Point Depression
\[
\Delta T_f = T_f^0 – T_f
\]
Since water freezes at \(0^\circ C\),
\[
\Delta T_f = 0 – (-0.24) = 0.24^\circ C
\]

Step 2: Use Freezing Point Depression Formula
\[
\Delta T_f = i \cdot K_f \cdot m
\]
where:
\( i \) = Van’t Hoff factor
\( K_f = 1.86^\circ C \cdot \text{kg}^{-1} \)
\( m \) = Molality

Step 3: Calculate Molality (m)
Molality is given by:
\[
m = \frac{\text{moles of solute}}{\text{kg of solvent}}
\]

Moles of KCl:
\[
\frac{0.5}{74.5} = 0.00671 \text{ moles}
\]

Mass of water in kg:
\[
\frac{100}{1000} = 0.1 \text{ kg}
\]

\[
m = \frac{0.00671}{0.1} = 0.0671 \text{ mol/kg}
\]

Step 4:
\[
i = \frac{\Delta T_f}{K_f \cdot m}
\]

\[
i = \frac{0.24}{1.86 \times 0.0671}
\]

\[
i = \frac{0.24}{0.1248} = 1.92
\]

Step 5: Calculate Percentage Ionization
For KCl dissociation:
\[
KCl \rightarrow K^+ + Cl^-
\]

Using the Van’t Hoff factor equation:
\[
i = 1 + \alpha (n – 1)
\]
Since KCl dissociates into 2 ions, \( n = 2 \):
\[
1.92 = 1 + \alpha (2 – 1)
\]

\[
1.92 = 1 + \alpha
\]

\[
\alpha = 1.92 – 1 = 0.92
\]

\[
\alpha \times 100 = 92\%
\]

Percentage Ionization of KCl = 92%

Van’t Hoff Factors in case of Association

When solute is completely associate in a solution, Van’t Hoff Factor can be calculated by the formula as:

$$i = \frac{\text{No. of moles after association}}{No.of moles taken}$$

Let n simple molecules of the solute A associate to form the associated molecule A_n. Hence, nA ⇌ A_n Now we shall proceed with the following data.

Where ‘α’ is the degree of association and $$\alpha = \frac{\text{No. of moles associated}}{\text {No. of moles of solute taken}}$$ Now $$i =\frac{1 – \alpha + \frac{\alpha}{n}}{1}$$ or $$\alpha =(1 – i)\cdot \frac{n}{n-1}$$ and $$i = \frac{(M_o – M_c)}{M_o}\cdot \frac{n}{(n – 1)}$$

Modified Expression of Colligative Properties for any type of solution

1. For lowering of vapour pressure $$\frac{P^o_B – P_B}{P^o_B} = i \cdot{x_A}$$ Where X_A is the mole fraction of solute.
2. For Osmotic pressure: $$\pi = i \cdot C \cdot R \cdot T$$
3. For Elevation in boiling point:$$\Delta{T_b} = i \cdot {K_b}\cdot{m}$$
4. For Depression in freezing point: $$\Delta{T_f} = i \cdot{K_f}\cdot{m}$$