Van’t Hoff Factor Shortcut for Percentage Dissociation

 

❓ Concept Question

How do we calculate percentage dissociation using van’t Hoff factor?

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🖼 Concept Image

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✍️ Short Concept

This concept is based on:

👉 Dissociation of electrolytes
👉 Number of particles formed
👉 van’t Hoff factor relation.

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🔷 Step 1 — Salt Dissociation 💯

Given salt:

$$𝑀{𝑋}_3$$

Dissociation:

$$𝑀{𝑋}_3\to {𝑀}^{3+}+3{𝑋}^{-}$$

Total particles formed:

$$𝑛=4$$

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🔷 Step 2 — van’t Hoff Relation

For dissociation:

$$𝑖=1+(𝑛-1)𝛼$$

Where:

• $𝑖$ = van’t Hoff factor
• $𝑛$ = total ions formed
• $𝛼$ = degree of dissociation

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🔷 Step 3 — Substitute Values

Given:

$$𝑖=2$$
$$𝑛=4$$

So:

$$2=1+(4-1)𝛼$$
$$2=1+3𝛼$$
$$𝛼=\frac{1}{3}$$

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🔷 Step 4 — Percentage Dissociation

$$\mathrm{\%}\text{ dissociation}=𝛼\times 100$$
$$=\frac{1}{3}\times 100$$
$$\approx 33\mathrm{\%}$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ $𝑛=3$ assume kar lena

❌ $𝛼$ ko directly percentage likh dena

❌ van’t Hoff formula bhool jaana

Remember:

$$𝑛=\text{total particles after dissociation}$$

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✅ Final Takeaway

For dissociation problems:

$$\boxed{{\displaystyle 𝑖=1+(𝑛-1)𝛼}}$$

Always:

$$\text{Count particles carefully first}$$

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⭐ Golden JEE Insight

If:

$$𝑖>1$$

👉 Dissociation happening

If:

$$𝑖<1$$

👉 Association happening