Question bank/JEE Main/Chemistry

JEE MainChemistryModeratePhysical Chemistry

The van't Hoff factor for 0.1 M Ba(NO3)2 solution is 2.74. The
degree of dissociation is

Show answer and solution
Van't Hoff factor, i=1+(n−1)αi = 1 + (n - 1)\alpha
Here n = 3
So, 2.74+1+(3−1)α2.74 + 1 + (3 - 1)\alpha
So, α=0.87 i.e. 87%\alpha = 0.87\ i.e.\ 87\%
okWUV gkWQ esa] i=1+(n−1)αi = 1 + (n - 1)\alpha;gk¡ n = 3

Source: Grass-G / CH-27 Chemistry Paper 2 15 Jan.docx

More JEE Main questionsChapter practice

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.

Scan a doubtView plans