JEE Advanced 2026 — Paper 2 — CHEM-III solution

JEE Advanced2026Paper 2CHEM-III

In a solvent $S$ , a compound $B$ is partially dissociated into $C$ and $D$ as given below:

$B ⇌ 2 C + 2 D$

$B$ , $C$ and $D$ are non-volatile in nature. The molar mass of $B$ is $10$ times the molar mass of $S$ . The standard boiling point and the standard enthalpy of vaporization of $S$ are $400$ K and $10 R$ J mol $^{- 1}$ , respectively ( $R$ is the gas constant in J K $^{- 1}$ mol $^{- 1}$ ). A solution of $B$ in $S$ with an initial concentration of $B$ as $0.25%$ (mass/mass) has a boiling point of $408$ K at $1$ bar pressure. In this solution, the mole percent of $B$ that has been dissociated is _____.

Your answer

Solution

van't Hoff factor. One mole of B gives 4 moles of particles, so $i = 1 + (4 - 1) α = 1 + 3 α$ .

Ebullioscopic constant of S.

$K_{b} = \frac{R ( T _{b}^{\circ} ) ^{2} M _{S}}{1000 Δ H _{vap}} = \frac{R ( 400 ) ^{2} M _{S}}{1000 \cdot 10 R} = \frac{160000 M _{S}}{10000} = 16 M_{S}$ K kg mol $^{- 1}$ .

Molality of B. $0.25%$ (m/m): in 100 g solution there is 0.25 g B in 99.75 g S.

$m = \frac{0.25/ M _{B}}{99.75/1000} = \frac{1000}{399 M _{B}}$ .

Apply $Δ T_{b} = i K_{b} m$ . $Δ T_{b} = 408 - 400 = 8$ K. Substituting $M_{B} = 10 M_{S}$ :

$8 = (1 + 3 α) (16 M_{S}) \frac{1000}{399 \cdot 10 M _{S}} = (1 + 3 α) \cdot \frac{16000}{3990}$ .

$1 + 3 α = \frac{8 \cdot 3990}{16000} = \frac{31920}{16000} = 1.995$ .

$3 α = 0.995 \Rightarrow α \approx 0.33166 \dots$

$∴$ Mole percent dissociated $\approx 33.17%$ .

← Back to full paper

Practice more CHEM-III

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →