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Van't Hoff factor
Appears in
Concepts tested here
- Number of ions on dissociation
- vant-hoff-factor-baoh2
All Questions
2016 Phase II 1 question
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The van't Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is
One formula unit gives three ions.
$Ba(OH)_2$ is a strong electrolyte, so 100% dissociation occurs in solution: $Ba(OH)_2 \rightarrow Ba^{2+}(aq) + 2OH^-(aq)$
Van't Hoff factor = total number of ions present in solution, i = 3
2015 AIPMT-I 1 question
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Which one of the following electrolytes has the same value of van't Hoff's factor (i) as that of $Al_2(SO_4)_3$ (if all are 100% ionized)?$Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$, i = 5
$K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$, i = 5
($K_2SO_4$: i = 3; $K_3[Fe(CN)_6]$: i = 4; $Al(NO_3)_3$: i = 4)