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NVAEasyJEE 2025Abnormal Molar Mass & van't Hoff Factor

JEE Chemistry 2025 Question with Solution

The percentage dissociation of a salt (MX3\text{MX}_3) solution at a given temperature (van't Hoff factor i=2i = 2) is _____ % (Nearest integer)

Answer

Correct answer:33

Step-by-step solution

Standard Method

Given: van't Hoff factor is i=2i = 2 for the salt MX3\text{MX}_3.

Find: percentage dissociation of the salt.

The salt dissociates as

MX3→M3++3X−\text{MX}_3 \rightarrow \text{M}^{3+} + 3\text{X}^-

So, one formula unit gives a total of 44 particles.

Use the relation

i=1+α(n−1)i = 1 + \alpha (n - 1)

where α\alpha is the degree of dissociation and n=4n = 4.

Substituting the given value:

2=1+α(4−1)2 = 1 + \alpha(4 - 1) 2=1+3α2 = 1 + 3\alpha 3α=13\alpha = 1 α=13\alpha = \frac{1}{3}

Now, percentage dissociation is

Percentage dissociation=α×100=13×100=33.33%\text{Percentage dissociation} = \alpha \times 100 = \frac{1}{3} \times 100 = 33.33\%

To the nearest integer, the percentage dissociation is 33%33\%.

Therefore, the required numerical answer is 33.

The solution also states the final answer as 33.

Detailed Relation Using van't Hoff Factor

Given: the solute is MX3\text{MX}_3 and the observed van't Hoff factor is i=2i = 2.

Find: the percentage dissociation.

If MX3\text{MX}_3 dissociates completely, it forms one ion of M3+\text{M}^{3+} and three ions of X−\text{X}^-, so the total number of ions formed per formula unit is n=4n = 4.

For dissociation,

i=1+α(n−1)i = 1 + \alpha(n - 1)

Putting n=4n = 4 and i=2i = 2,

2=1+α(3)2 = 1 + \alpha(3) 2=1+3α2 = 1 + 3\alpha α=2−13=13\alpha = \frac{2 - 1}{3} = \frac{1}{3}

Convert this into percentage:

α×100=13×100=33.33%\alpha \times 100 = \frac{1}{3} \times 100 = 33.33\%

Nearest integer value is 3333.

Therefore, the percentage dissociation is 33%33\%.

Common mistakes

  • Using the wrong value of nn. For MX3\text{MX}_3, the total ions formed are 44, not 33. Count both the cation and the anions before applying i=1+α(n−1)i = 1 + \alpha(n-1).

  • Taking van't Hoff factor ii directly as the degree of dissociation. The factor ii measures particles in solution, whereas dissociation is represented by α\alpha. First use the relation between ii and α\alpha.

  • Forgetting to convert the degree of dissociation into percentage. After getting α=13\alpha = \frac{1}{3}, multiply by 100100 to obtain the percentage dissociation.

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