Question

A plot given shows P – T curves (where P is the pressure and T is the temperature) for two solvents X and Y, and isomolar solutions of NaCl in these solvents. NaCl completely dissociates in both the solvents. In addition to an equal number of moles of non-volatile solute S in equal amount (in Kg) of these solvents, the elevation of the boiling point of solvent X is three times that of solvent. 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:

Answer 1

Hint: The elevation in the boiling point is equal to the product of ionization of solute, molality, and molal elevation constant. The formula is $\mathrm{\Delta }{T}_b=i\text{ x m x }{\text{K}}_b$. The value of i can be calculated by the reaction $2(S)\to S_2$.

Complete answer: In the graph, there are four lines in which the first 2 lines are used for solvent X and the last two lines are used for solvent Y.
Line 2 is at 362 K and line 1 is at 360 K, so we can calculate the elevation in the boiling point as:
 $\mathrm{\Delta }{T}_{b(X)}=362-360=2$
Line 4 is at 368 K and line 3 is at 367 K, so we can calculate the elevation in the boiling point as:
$\mathrm{\Delta }{T}_{b(Y)}=368-367=1$
According to the formula, we can write for X and Y as:
$\mathrm{\Delta }{T}_{b(X)}=i\text{ x }{\text{m}}_{NaCl}\text{ x }{\text{K}}_{b(X)}$
$\mathrm{\Delta }{T}_{b(Y)}=i\text{ x }{\text{m}}_{NaCl}\text{ x }{\text{K}}_{b(Y)}$
When we divide the above equations, we get:
${\displaystyle \frac{K_{b(X)}}{K_{b(Y)}}}=2$
Since the dimerization takes place, we can write the equation as:
$2(S)\to S_2$
After the equilibrium is attained, the concentration of S will be $1-\alpha$ and the value of $S_2$ will be ${\displaystyle \frac{\alpha }{2}}$
So, the value of i, will be:
$i=(1-{\displaystyle \frac{\alpha }{2}})$
Now, putting these values in the elevation in boiling point equation, we get:
$\mathrm{\Delta }{T}_{b(X)}=(1-{\displaystyle \frac{{\alpha }_1}{2}})\text{ x }{\text{m}}_{NaCl}\text{ x }{\text{K}}_{b(X)}$
$\mathrm{\Delta }{T}_{b(Y)}=(1-{\displaystyle \frac{{\alpha }_2}{2}})\text{ x }{\text{m}}_{NaCl}\text{ x }{\text{K}}_{b(Y)}$
In the question, it is given that:
$\mathrm{\Delta }{T}_{b(X)}=3\text{ x }\mathrm{\Delta }{T}_{b(Y)}$
Combining, all these we can write:
$(1-{\displaystyle \frac{{\alpha }_1}{2}})\text{ x }{\text{K}}_{b(X)}=3\text{ x }(1-{\displaystyle \frac{{\alpha }_2}{2}})\text{ x }{\text{K}}_{b(X)}$
$2\text{ x }(1-{\displaystyle \frac{{\alpha }_1}{2}})=3\text{ x }(1-{\displaystyle \frac{{\alpha }_2}{2}})$
$a_2$ = 0.7 and $a_1$ = 0.05.
The degree of dimerization of solvent X is 0.05.

Note: Don’t get confused between the formulas $\mathrm{\Delta }{T}_b=i\text{ x m x }{\text{K}}_b$ and $\mathrm{\Delta }{T}_b=\text{ m x }{\text{K}}_b$, the former is used when there is any electrolyte or ionic compound is present in the solution and the latter is used when covalent compound is present.