JEE Advanced 2018 · Paper 1 · Chemistry Q14

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ChemistryNumerical+3Question 32 of 54

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 of 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 ______.

Vapour pressure (mmHg) against temperature (K). At 760 mmHg: curve 1 (solvent X) reaches it at 360 K, curve 2 (NaCl solution in X) at 362 K, curve 3 (solvent Y) at 367 K and curve 4 (NaCl solution in Y) at 368 K.

Your answer

Answer and worked solution

Official answer: 0.05

Ebullioscopic constants. At 760 mmHg, solvent X boils at 360 K and its NaCl solution at 362 K, so $\Delta T_b=2$ K. For Y the values are 367 K and 368 K, so $\Delta T_b=1$ K.

The NaCl solutions are isomolal and $i=2$ in both, so $K_b(X)=2K_b(Y)$ .

Solute S. It has the same molality in both solvents. For dimerisation with degree $\alpha$ , $i=1-\alpha/2$ .
$\Delta T_X=3\Delta T_Y\implies K_b(X)\,i_X=3K_b(Y)\,i_Y\implies2i_X=3i_Y.$

Solve. $i_Y=1-0.35=0.65$ , so $i_X=0.975=1-\alpha_X/2$ , giving $\alpha_X=0.05$ .

Answer: 0.05

Topics:Colligative properties van't Hoff factor

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JEE Advanced 2018 · Paper 1, Chemistry question 14