JEE MAIN - Chemistry (2024 - 1st February Morning Shift - No. 11)

We have three aqueous solutions of $\mathrm{NaCl}$ labelled as ' $\mathrm{A}$ ', ' $\mathrm{B}$ ' and ' $\mathrm{C}$ ' with concentration $0.1 \mathrm{M}$, $0.01 \mathrm{M}$ and $0.001 \mathrm{M}$, respectively. The value of van 't Hoff factor(i) for these solutions will be in the order :
$\mathrm{i}_{\mathrm{A}}<\mathrm{i}_{\mathrm{C}}<\mathrm{i}_{\mathrm{B}}$
$\mathrm{i}_{\mathrm{A}}<\mathrm{i}_{\mathrm{B}}<\mathrm{i}_{\mathrm{C}}$
$\mathrm{i}_{\mathrm{A}}>\mathrm{i}_{\mathrm{B}}>\mathrm{i}_{\mathrm{C}}$
$\mathrm{i}_{\mathrm{A}}=\mathrm{i}_{\mathrm{B}}=\mathrm{i}_{\mathrm{C}}$

Explanation

Salt Values of i (for different conc. of a Salt)
NaCl 0.1 M 0.01 M 0.001 M
1.87 1.94 1.94

The van 't Hoff factor (i) is used to describe the number of particles a solute formula unit produces in a solution. For an electrolyte like $\mathrm{NaCl}$, which dissociates completely in very dilute solutions, the theoretical value of $i$ is approximately 2, since $\mathrm{NaCl}$ dissociates into $\mathrm{Na}^+$ and $\mathrm{Cl}^-$ ions.

In real scenarios, as the concentration of the solution decreases (making the solution more dilute), the interaction between the ions decreases, allowing more complete dissociation. Therefore, for practical purposes, the van 't Hoff factor $i$ approaches its theoretical maximum value as concentration decreases. Thus, for $\mathrm{NaCl}$ solutions of concentrations $0.1 \mathrm{M}$, $0.01 \mathrm{M}$, and $0.001 \mathrm{M}$, the fact that $\mathrm{NaCl}$ dissociates more completely in more dilute solutions implies that $i$ increases with decreasing concentration.

Hence, the order of $i$ based on the concentration would be $\mathrm{i}_{\mathrm{A}} < \mathrm{i}_{\mathrm{B}} < \mathrm{i}_{\mathrm{C}}$, as concentration $0.1 \mathrm{M} > 0.01 \mathrm{M} > 0.001 \mathrm{M}$, respectively. Thus, option B correctly describes the order of the van 't Hoff factors for these solutions.

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