The plot given below shows P-T curves (where P is the pressure and T is the temperature) for two sol

Q: Why do we look at the intersection with the 760 mmHg line?

The normal boiling point of a liquid is the temperature at which its vapor pressure equals the atmospheric pressure, which is typically 1 atm or 760 mmHg.

Q: How did we get $i = 1 - \alpha/2$ for dimerization?

For a dimerization reaction $2S \rightleftharpoons S_2$, if we start with 1 mole of S, $1-\alpha$ moles remain unreacted, and $\alpha/2$ moles of $S_2$ are formed. The total number of moles is $1 - \alpha + \alpha/2 = 1 - \alpha/2$.

Q: Why did the molality $m'$ cancel out in the second part?

The problem states that an equal number of moles of solute S were added to equal amounts (in kg) of both solvents. Since molality is moles of solute per kg of solvent, the molality $m'$ is identical for both solutions.

Q: What does the degree of dimerization $\alpha = 0.05$ mean physically?

It means that only 5% of the solute S molecules in solvent X undergo dimerization to form $S_2$ molecules, while the remaining 95% stay as individual S molecules.

Question

The plot given below shows P–T curves (where P is the pressure and T is the temperature) for two solvents X and Y and isomolal solutions of NaCl in these solvents. NaCl completely dissociates in both the solvents.

On addition of equal number of moles a non-volatile solute S in equal amount (in kg) of these solvents, the elevation of boiling point of solvent X is three times that of solvent Y. Solute S is known to undergo dimerization in these solvents. If the degree of dimerization is 0.7 in solvent Y, the degree of dimerization in solvent X is ___.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are in a chemistry lab, carefully observing the boiling points of different solutions. The graph provided is our window into this physical reality. It plots vapor pressure against temperature.

We know that a liquid boils when its vapor pressure equals the external atmospheric pressure, which is standardly $760 	ext{ mmHg}$. By tracing the horizontal line at $760 	ext{ mmHg}$, we can pinpoint the exact boiling temperatures.

For pure solvent X, the curve intersects this line at $360 	ext{ K}$. When we add NaCl to it, the boiling point elevates to $362 	ext{ K}$. This gives us an elevation in boiling point, $\Delta T_{bx} = 2 	ext{ K}$.

Similarly, for pure solvent Y, the boiling point is $367 	ext{ K}$, and its NaCl solution boils at $368 	ext{ K}$. Thus, the elevation for solvent Y is $\Delta T_{by} = 1 	ext{ K}$.

### The Master Equation

Now, let's bring in our theoretical tool: the formula for the elevation in boiling point, $\Delta T_b = i \cdot m \cdot K_b$.

Here, $i$ is the van't Hoff factor, $m$ is the molality, and $K_b$ is the ebullioscopic constant of the solvent. Since NaCl is a strong electrolyte that completely dissociates into two ions ($	ext{Na}^+$ and $	ext{Cl}^-$), its van't Hoff factor $i$ is $2$.

The problem states that the NaCl solutions in both solvents are isomolal, meaning they have the exact same molality, $m$. If we take the ratio of their boiling point elevations, the $i$ and $m$ terms beautifully cancel out!

We get $\frac{\Delta T_{bx}}{\Delta T_{by}} = \frac{K_{bx}}{K_{by}}$. Substituting our values from the graph, we find that $\frac{K_{bx}}{K_{by}} = \frac{2}{1} = 2$. This tells us that the ebullioscopic constant of solvent X is exactly twice that of solvent Y.

### The Dimerization Challenge

Next, the problem introduces a new, non-volatile solute S. This solute is tricky—it undergoes dimerization, meaning two molecules pair up to form a single entity ($2S ightleftharpoons S_2$).

For such a process, the van't Hoff factor is given by $i = 1 - \frac{\alpha}{2}$, where $\alpha$ is the degree of dimerization. We are told that equal moles of S are added to equal masses of both solvents, which means the new molality $m'$ is identical for both new solutions.

The crucial condition given is that the elevation in boiling point for solvent X is three times that of solvent Y: $\Delta T'_{bx} = 3 \Delta T'_{by}$.

Substituting our formula, we get $i_x \cdot m' \cdot K_{bx} = 3 \cdot i_y \cdot m' \cdot K_{by}$. The $m'$ cancels out, leaving us with a clean relationship between the van't Hoff factors and the ebullioscopic constants.

### Final Calculation

Let's plug in everything we know. We substitute $K_{bx} = 2 K_{by}$ and the given degree of dimerization for solvent Y, $\alpha_y = 0.7$.

Our equation becomes $2 \left(1 - \frac{\alpha_x}{2}ight) = 3 \left(1 - \frac{0.7}{2}ight)$.

The right side simplifies to $3 	imes (1 - 0.35) = 3 	imes 0.65 = 1.95$.

So, we have $2 - \alpha_x = 1.95$. Solving for $\alpha_x$, we get $\alpha_x = 2 - 1.95 = 0.05$.

**The degree of dimerization of solute S in solvent X is 0.05.** This means only 5% of the molecules dimerize in solvent X, compared to 70% in solvent Y!

Enter Numerical Value:

Visualized Solution

\text{Analyzing the Graph}

\text{Elevation in Boiling Point}

\text{Formula for } \Delta T_b

\text{Ratio of Ebullioscopic Constants}

\text{Calculating } K_{bx} / K_{by}

\text{Dimerization of Solute S}

\text{Condition for Solute S}

\text{Substituting Values}

\text{Final Calculation}

\text{The Way Forward}

The Sigma Insight: Colligative Properties

Analyzing the Setup

The Master Equation

The Dimerization Challenge

Final Calculation

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