Solutions · Question 1

JEE Advanced 2026 Paper 2NumericalHard📈 Topic in 33% papers+4

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

Answer

31.60 to 33.33

Solution

Kb= RTb2MS1000 vapH= R (400)2MS1000 10R=16MS K kg mol-1 (MS in g mol-1)

0.25 g B (MB=10MS) in 99.75 g S: m= 0.25/(10MS)0.09975= 0.2506MS

Tb=iKbm 8=i 16MS 0.2506MS=4.01 i i=1.995

For B 2C+2D: i=1- +4 =1+3 = 0.9953=0.332

Mole percent dissociated 33.2

Answer: 33.17 (accepted 31.60 to 33.33)

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