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NVAEasyJEE 2023Abnormal Molar Mass & van't Hoff Factor

JEE Chemistry 2023 Question with Solution

Consider the following pairs of solution which will be isotonic at the same temperature. The number of pairs of solutions is/are_____

A. 1 M1 \, \text{M} aq. NaCl and 2 M2 \, \text{M} aq. Urea

B. 1 M1 \, \text{M} aq. CaCl2_2 and 1.5 M1.5 \, \text{M} aq. KCl

C. 1.5 M1.5 \, \text{M} aq. AlCl3_3 and 2 M2 \, \text{M} aq. Na2_2SO4_4

D. 2.5 M2.5 \, \text{M} aq. KCl and 1 M1 \, \text{M} aq. Al2_2(SO4_4)3_3

Answer

Correct answer:4

Step-by-step solution

Standard Method

Given: Pairs of solutions are to be checked for isotonic behavior at the same temperature.

Find: The number of pairs that are isotonic.

Isotonic solutions have the same osmotic pressure. For dilute solutions at the same temperature,

π=iCRT\pi = iCRT

So, isotonic pairs must have the same value of iCiC, that is, the same total concentration of solute particles.

For A:

NaCl→Na++Cl−\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-

So, i=2i = 2 and

1×2=21 \times 2 = 2

For urea, there is no dissociation, so i=1i = 1 and

2×1=22 \times 1 = 2

Both have the same particle concentration, so A is isotonic.

For B:

CaCl2→Ca2++2Cl−\text{CaCl}_2 \rightarrow \text{Ca}^{2+} + 2\text{Cl}^-

So, i=3i = 3 and

1×3=31 \times 3 = 3

Also,

KCl→K++Cl−\text{KCl} \rightarrow \text{K}^+ + \text{Cl}^-

So, i=2i = 2 and

1.5×2=31.5 \times 2 = 3

Both have the same particle concentration, so B is isotonic.

For C:

AlCl3→Al3++3Cl−\text{AlCl}_3 \rightarrow \text{Al}^{3+} + 3\text{Cl}^-

So, i=4i = 4 and

1.5×4=61.5 \times 4 = 6

Also,

Na2SO4→2Na++SO42−\text{Na}_2\text{SO}_4 \rightarrow 2\text{Na}^+ + \text{SO}_4^{2-}

So, i=3i = 3 and

2×3=62 \times 3 = 6

Both have the same particle concentration, so C is isotonic.

For D:

KCl→K++Cl−\text{KCl} \rightarrow \text{K}^+ + \text{Cl}^-

So, i=2i = 2 and

2.5×2=52.5 \times 2 = 5

Also,

Al2(SO4)3→2Al3++3SO42−\text{Al}_2(\text{SO}_4)_3 \rightarrow 2\text{Al}^{3+} + 3\text{SO}_4^{2-}

So, i=5i = 5 and

1×5=51 \times 5 = 5

Both have the same particle concentration, so D is isotonic.

All four pairs satisfy the isotonic condition.

Therefore, the number of isotonic pairs is 44.

Compare particle concentration directly

Given: Each pair must be checked for equality of total ion concentration.

Find: How many pairs are isotonic.

A quick method is to multiply molarity by the number of particles formed after dissociation.

A: 1×2=2,2×1=2B: 1×3=3,1.5×2=3C: 1.5×4=6,2×3=6D: 2.5×2=5,1×5=5\begin{aligned} \text{A: } &1 \times 2 = 2, \quad 2 \times 1 = 2 \\ \text{B: } &1 \times 3 = 3, \quad 1.5 \times 2 = 3 \\ \text{C: } &1.5 \times 4 = 6, \quad 2 \times 3 = 6 \\ \text{D: } &2.5 \times 2 = 5, \quad 1 \times 5 = 5 \end{aligned}

Each pair has equal particle concentration, so every pair is isotonic.

Therefore, the required number is 44.

Common mistakes

  • Treating isotonicity as depending only on molarity is incorrect because osmotic pressure depends on the total number of solute particles. Always compare iCiC, not just CC.

  • Ignoring dissociation of ionic solutes gives wrong particle counts. For salts such as NaCl, CaCl2_2, and AlCl3_3, first count how many ions each formula unit produces.

  • Assuming urea also dissociates is wrong because urea is a non-electrolyte in aqueous solution. Use i=1i = 1 for urea.

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