the van't Hoff factor for a dilute aqueous solution of a strong electrolyte barium Hydroxide is
Answer (1)
8th Jan, 2020
Dear Student
as it is stated that it is the ratio of number of particles after association to the number of particles before association.
We all know that vant Hoff factor is equal to ions dissassociated .So for Ba(oh)2
Vant Hoff factor will be 3.As it will dissassociate in Ba2+ and 2OH-
Thanks
All the very best
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