Calculate Van't Hoff Factor with Degree of Dissociation | JEE & NEET - CHEMCA

How to Calculate Van't Hoff Factor from Degree of Dissociation (α)

Published by Abhishek Sengar | CHEMCA India

In the Physical Chemistry chapter of Solutions and Colligative Properties, dealing with electrolytes means you have to account for the Van't Hoff factor (i). While strong electrolytes fully dissociate, weak electrolytes only partially break down. This partial breakdown is measured by the Degree of Dissociation (α).

If you are aiming for JEE Main, JEE Advanced, or NEET, knowing the exact formula connecting i and α is non-negotiable. Let's break down a numerical problem to understand its application.

Video Tutorial: Step-by-Step Calculation

Watch Abhishek Sengar sir from CHEMCA solve a classic question: "What will be the Van't Hoff factor of Mg₃(PO₄)₂ if α is 0.6?"

The Master Formula

To find the Van't Hoff factor for a solute undergoing dissociation, we use the following standard relation:

α = (i - 1) / (n - 1)

• α (Alpha): Degree of dissociation (given as 0.6 in our example).
• i: Van't Hoff factor (what we need to find).
• n: Total number of ions produced by the dissociation of one molecule of the electrolyte.

Pro-Tip from Sir: If the problem states the solute is an electrolyte but doesn't mention the value of α, you must assume it is a strong electrolyte with 100% dissociation. In that case, α = 1, which simplifies the formula directly to i = n.

Solved Example: Magnesium Phosphate

Let's solve the exact question from the video: Find i for Magnesium Phosphate Mg₃(PO₄)₂ given α = 0.6.

1. Find the number of ions (n):
   Ionization equation: Mg₃(PO₄)₂ → 3Mg²⁺ + 2PO₄^3-
   Total ions: n = 3 + 2 = 5
2. Apply the formula:
   α = (i - 1) / (n - 1)
   0.6 = (i - 1) / (5 - 1)
3. Solve for i:
   0.6 = (i - 1) / 4
   0.6 × 4 = i - 1
   2.4 = i - 1
   i = 3.4

Notice that because dissociation is partial (60%), the Van't Hoff factor (3.4) is greater than 1, but less than the theoretical maximum of 5.

Practice Questions for JEE & NEET

Test your understanding with these exam-level questions. Click the button to verify your calculations!

Question 1: Calculate the Van't Hoff factor for an aqueous solution of Aluminium Sulphate Al₂(SO₄)₃ if it is 80% dissociated.

Step-by-step Solution:

• Given: α = 80% = 0.8
• Ionization: Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄^2-
• Number of ions (n) = 2 + 3 = 5
• Formula: α = (i - 1) / (n - 1)
• 0.8 = (i - 1) / (5 - 1)
• 0.8 = (i - 1) / 4
• 3.2 = i - 1
• Answer: i = 4.2

Question 2: An electrolyte A₂B₃ is dissolved in water. If no information about its degree of dissociation is given, what will be the value of its Van't Hoff factor (i) used in colligative property formulas?

Answer: 5

Reasoning: As Abhishek Sir pointed out, if α is not mentioned, we assume the compound is a strong electrolyte undergoing 100% dissociation. Therefore, α = 1.

The compound A₂B₃ breaks into 2 A ions and 3 B ions. Thus, n = 2 + 3 = 5.

When α = 1, the formula α = (i - 1) / (n - 1) simplifies to i = n. Therefore, i = 5.

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