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 10R J mol^ -1 , respec

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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 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 ____.

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Correct answer: A. A

The elevation in boiling point is given by T_b = i K_b m. The ebullioscopic constant K_b is calculated as: K_b = R (T_b^ )^2 M_S 1000 H_ vap Substituting the given values (T_b^ = 400 K, H_ vap = 10R J mol^ -1 ): K_b = R (400)^2 M_S 1000 (10R) = 16 M_S For a dilute solution with 0.25\% (mass/mass) concentration, the mass of solute w_B = 0.25 g and the mass of solvent w_S 100 g. The molality m is: m = w_B 1000 M_B w_S = 0.25 1000 10 M_S 100 = 0.25 M_S Given T_b = 408 - 400 = 8 K, we substitute into the boiling point elevation formula: 8 = i (16 M_S) ( 0.25 M_S ) 8 = 4i i = 2 The dissociation reaction is B 2 C + 2 D . The van't Hoff factor i is related to the degree of dissociation by: i = 1 + (n - 1) Here, n = 4 (since 1 molecule of B yields 4 particles). 2 = 1 + (4 - 1) 3 = 1 = 1 3 The mole percent of B that has been dissociated is 100 = 33.33\%. Answer: 33.33

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