1. Use the given $P$–$T$ graph information for isomolal $\mathrm{NaCl}$ solutions

For boiling point elevation, $\Delta T_b = i K_b m$ where:

• $i$ = van’t Hoff factor
• $K_b$ = ebullioscopic constant of the solvent
• $m$ = molality

For $\mathrm{NaCl}$, complete dissociation is given, so $i_{\mathrm{NaCl}} = 2$

Since the solutions are isomolal, $m$ is same for both solvents. Hence, $\Delta T_b \propto K_b$ for these NaCl solutions.

From the graph, the boiling point elevation for the isomolal NaCl solution in solvent $X$ is twice that in solvent $Y$. Therefore, $\frac{K_b(X)}{K_b(Y)}=2$

2. Now consider solute $S$

Equal moles of $S$ are added to equal kg of both solvents, so molality is again same in both cases.

Given: $\Delta T_b(X) = 3\,\Delta T_b(Y)$

Using $\Delta T_b = i K_b m$ we get $i_X K_b(X) = 3\, i_Y K_b(Y)$

Substitute $K_b(X)=2K_b(Y)$: $i_X \cdot 2K_b(Y) = 3 i_Y K_b(Y)$ $2 i_X = 3 i_Y$ $i_X = \frac{3}{2} i_Y$

3. Relate van’t Hoff factor to dimerization

For dimerization: $2A \rightarrow A_2$ If degree of dimerization is $\alpha$, then starting from 1 mole:

• moles of monomer left $=1-\alpha$
• moles of dimer formed $=\alpha/2$

Total moles after association: $1-\alpha+\frac{\alpha}{2}=1-\frac{\alpha}{2}$ So van’t Hoff factor is $i = 1-\frac{\alpha}{2}$

4. Use the given dimerization in solvent $Y$

Given: $\alpha_Y = 0.7$ Thus, $i_Y = 1-\frac{0.7}{2}=1-0.35=0.65$

Then, $i_X = \frac{3}{2}\times 0.65 = 0.975$

Now, $i_X = 1-\frac{\alpha_X}{2}$ So, $0.975 = 1-\frac{\alpha_X}{2}$ $\frac{\alpha_X}{2}=0.025$ $\alpha_X = 0.05$

5. Final answer

The degree of dimerization in solvent $X$ is $\boxed{0.05}$