JEE Advanced 2026 · Paper 2 · Chemistry Q12

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ChemistryNumerical+4Question 48 of 54

In a solvent S, a compound B is partially dissociated into C and D as given below:
$\ce{B <=> 2C + 2D}$
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 $10R$ 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

Answer and worked solution

Official answer: 31.6

Ebullioscopic constant. With $M_S$ in g mol⁻¹,
$K_b=\frac{RT_b^2M_S}{1000\,\Delta H_{\text{vap}}}=\frac{R(400)^2M_S}{1000\times10R}=16M_S\ \text{K kg mol}^{-1}.$

Molality. The solution has 0.25 g of B ( $M_B=10M_S$ ) in 99.75 g of S:
$m=\frac{0.25/(10M_S)}{0.09975}=\frac{0.2506}{M_S}.$
So $K_bm=4.01$ K.

Dissociation. For $\ce{B <=> 2C + 2D}$ , $i=1+3\alpha$ . Then
$8=i\times4.01\implies i=1.995\implies\alpha=0.332.$
About 33.2 % of B is dissociated. (Taking 100 g of solvent gives $i=2$ and $\alpha=\frac13$ .)

Topics:van't Hoff factor Colligative properties

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JEE Advanced 2026 · Paper 2, Chemistry question 12