We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentration 0.1 M, 0.01M & 0.001 M, respectively. The value of vant’ Haft factor (i) for these solutions will be in the order.

Question

$W$ g of a non-volatile electrolyte solid solute of molar mass $M\,\text{g mol}^{-1}$ when dissolved in $100\,\text{mL}$ water decreases vapour pressure of water from $640\,\text{mm Hg}$ to $600\,\text{mm Hg}$. If aqueous solution of the electrolyte boils at $375\,\text{K}$ and $K_b$ for water is $0.52\,\text{K kg mol}^{-1}$, then the mole fraction of the electrolyte $(x_2)$ in the solution can be expressed as ___.
(Given: density of water $=1\,\text{g mL}^{-1}$, boiling point of water $=373\,\text{K}$)

Answer 1

This problem requires determining the order of the van't Hoff factor ($i$) for three aqueous solutions of NaCl at varying concentrations: 0.1 M (A), 0.01 M (B), and 0.001 M (C).

Concept Used:

The van't Hoff factor ($i$) quantifies the extent of solute dissociation or association in a solution. For an electrolyte, it is defined as:

\[ i = \frac{\text{Actual number of particles in solution after dissociation}}{\text{Number of formula units initially dissolved in solution}} \]

For a dissociating electrolyte, the van't Hoff factor is linked to the degree of dissociation ($\alpha$) by:

\[ i = 1 + \alpha(n - 1) \]

where $n$ represents the number of ions generated from one formula unit of the solute.

Strong electrolytes like NaCl exhibit near-complete dissociation, but not 100% in practice, especially at higher concentrations. This deviation is attributed to inter-ionic attractions between ions. Higher concentrations lead to closer ion proximity, stronger attractions, and the formation of "ion pairs," thereby reducing the number of free-moving particles. Dilution increases the average ion separation, weakening inter-ionic forces and enhancing the degree of dissociation ($\alpha$). Consequently, the van't Hoff factor ($i$) increases with dilution and approaches the theoretical value ($n$) at infinite dilution.

Step-by-Step Solution:

Step 1: Analyze NaCl Dissociation.

Sodium chloride (NaCl), a strong electrolyte, dissociates in water into one sodium ion ($\text{Na}^+$) and one chloride ion ($\text{Cl}^-$).

\[ \text{NaCl}_{(aq)} \rightarrow \text{Na}^+_{(aq)} + \text{Cl}^-_{(aq)} \]

Each formula unit of NaCl yields two ions, making the theoretical maximum van't Hoff factor $n=2$.

Step 2: Relate van't Hoff Factor to Dissociation Degree for NaCl.

Substituting $n=2$ into the formula $i = 1 + \alpha(n - 1)$ yields:

\[ i = 1 + \alpha(2 - 1) = 1 + \alpha \]

This equation indicates that the van't Hoff factor ($i$) for NaCl is directly proportional to its degree of dissociation ($\alpha$).

Step 3: Analyze Concentration's Effect on Dissociation Degree.

The concentrations of the three solutions are:

Solution A: 0.1 M
Solution B: 0.01 M
Solution C: 0.001 M

The concentration order is: $ \text{Concentration (A)} > \text{Concentration (B)} > \text{Concentration (C)} $.

As electrolyte concentration decreases, the solution becomes more dilute. In dilute solutions, ions are farther apart, resulting in weaker inter-ionic forces and a greater degree of dissociation ($\alpha$).

Therefore, the order of the degree of dissociation will be the inverse of the concentration order:

\[ \alpha_C > \alpha_B > \alpha_A \]

Step 4: Determine the Order of the van't Hoff Factor.

Given that $i = 1 + \alpha$, the van't Hoff factor ($i$) will follow the same order as the degree of dissociation ($\alpha$).

Let $i_A$, $i_B$, and $i_C$ represent the van't Hoff factors for solutions A, B, and C, respectively. Since $ \alpha_C > \alpha_B > \alpha_A $, it follows that:

\[ i_C > i_B > i_A \]

Final Result:

The most dilute solution (C) will exhibit the highest degree of dissociation and consequently the largest van't Hoff factor. Conversely, the most concentrated solution (A) will have the lowest degree of dissociation and the smallest van't Hoff factor. The value of $i$ for all solutions will be below the theoretical maximum of 2, but will approach 2 as concentration decreases.

The order of the van't Hoff factor for these solutions is $i_A < i_B < i_C$.

We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentration 0.1 M, 0.01M & 0.001 M, respectively. The value of vant’ Haft factor (i) for these solutions will be in the order.

The Correct Option is A

Solution and Explanation

Concept Used:

Step-by-Step Solution:

Final Result:

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