To aqueous solution of NaI , increasing amounts of solid HgI 2 is added. The v | TestHub

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A system with greater molecular disorder is more probable. The disorder of molecules is reflected by the entropy of the system. A liquid vaporizes to form a more disordered gas. When a solute is present, there is an additional contribution to the entropy of the liquid due to increased randomness. As the entropy of the solution is higher than that of the pure liquid, there is a weaker tendency to form the gas. Thus, a non-volatile solute lowers the vapor pressure of a liquid, and hence results in a higher boiling point for the solution. Similarly, the greater randomness of the solution opposes the tendency to freeze. In consequence, a lower temperature must be reached to achieve equilibrium between the solid (frozen solvent) and the solution. Elevation of boiling point ( $Δ T_{b}$ ) and depression of freezing point ( $Δ T_{f}$ ) of a solution are colligative properties that depend only on the concentration of solute particles, not their identity. For dilute solutions, $Δ T_{b}$ and $Δ T_{f}$ are proportional to the molality of the solute in the solution.

$Δ T_{b} = K_{b} m$

$K_{b} = E b u ll i osco p i cco n s t an t = \frac{R T _{b}^{2} M}{1000Δ H _{v a p}}$

And $Δ T_{f} = K_{f} m$

$K_{f} = C ryosco p i cco n s t an t = \frac{R T _{f}^{2} M}{1000Δ H _{f u s}}$ (where $M$ is the molecular mass of the solvent).

The values of $K_{b}$ and $K_{f}$ depend on the properties of the solvent. For liquids, $\frac{Δ H _{v a p}}{T _{b}^{0}}$ is almost constant.

[According to Trouton's Rule, this constant for most unassociated liquids (those not having strong bonding like hydrogen bonding in the liquid state) is equal to $90 J / m o l$ .]

For solutes undergoing a change of molecular state in solution (ionization or association), the observed $Δ T$ values differ from those calculated using the above relations. In such situations, the relationships are modified as:

$Δ T_{b} = i K_{b} m$

$Δ T_{f} = i K_{f} m$

where $i$ is the Van't-Hoff factor, which is greater than unity for ionization and smaller than unity for association of the solute molecules.

To aqueous solution of NaI , increasing amounts of solid $HgI_{2}$ is added. The vapor pressure of the solution

Options:

Answer:

B

Solution:

$2 Na^{+} (aq) + 2 I^{-} (aq) + HgI_{2} (s) ⟶ Na^{+} (aq) + HgI_{4}^{2 -} (aq)$

The number of mole particle decreases from $4 (2 Na^{+} + 2 I^{-})$ to $3 (2 Na^{+} + HgI_{4}^{2 -})$ . Hence, the colligative property will decrease to, the vapour pressure will increase to a constant value until NaI is completely consumed.

Stream:JEESubject:ChemistryTopic:Liquid SolutionSubtopic:Colligative Properties

⏱ 2mℹ️ Source: QB

Chemistry - Liquid Solution Question with Solution | TestHub

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