Question

Van't hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is

Answer 1

Answer : The Van't Hoff factor for a dilute aqueous solution of the strong electrolyte barium hydroxide is, $i=3$

Explanation :

Van't Hoff factor is the ratio of number of particles after association to the number of particles before association.

$i=\frac{\text {observed colligative property}}{\text {Calculated Colligative property}}$

For 100% dissociation of a solute or for strong electrolyte, van't Hoff factor is equal to the number of ions produces from one molecule of the solute.

For barium hydroxide $(Ba(OH)_2)$ :

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^-$

After dissociation : the number of ions produced = 1 + 2 = 3

So, the Van't Hoff factor, $i=3$

Answer 2

1594448468494824008028