Chemistry Q72 – JEE-MAIN 2025

The percentage dissociation of a salt (

M X_{3}

) solution at given temperature (van't Hoff factor

i = 2

) is _______ %(Nearest integer)

Enter Calculated Numerical Value:

Clue 1 of 3: Dissociation of Electrolytes

Understand how a salt dissociates into ions in a solution.

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Step 1: Determine the number of ions (n) upon dissociation✦ Active

The salt $M X_{3}$ dissociates in solution as follows:

M X_{3} ⇌ M^{3 +} + 3 X^{-}

From the dissociation equation, 1 molecule of $M X_{3}$ produces 1 $M^{3 +}$ ion and 3 $X^{-}$ ions. Therefore, the total number of ions produced per formula unit, $n$ , is $1 + 3 = 4$ .

Step 2: Apply the van't Hoff factor formula○ Expand

The van't Hoff factor ( $i$ ) is related to the degree of dissociation ( $α$ ) and the number of ions ( $n$ ) by the formula:

i = 1 + (n - 1) α

Given $i = 2$ and from Step 1, $n = 4$ . Substitute these values into the formula:

2 = 1 + (4 - 1) α

Simplify the equation:

2 = 1 + 3 α

Subtract 1 from both sides:

1 = 3 α

Step 3: Calculate the percentage dissociation○ Expand

From Step 2, we found that $3 α = 1$ . Solve for $α$ :

α = \frac{1}{3}

To express this as a percentage dissociation, multiply by 100%:

Percentage dissociation = α \times 100 % = \frac{1}{3} \times 100 % = 33.33 . . . %

Rounding to the nearest integer, the percentage dissociation is $33 %$ .

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Chemistry Question 72 – JEE-MAIN 2025

Clue 1 of 3: Dissociation of Electrolytes

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