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The van't Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is
A
2
B
3
C
0
D
1
Explanation
One formula unit gives three ions.
Detailed Solution
$Ba(OH)_2$ 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
Van't Hoff factor = total number of ions present in solution, i = 3