Question

Which one of the following electrolytes has the same value of van't Hoff factor $\left(i\right)$ as that of $K_{4}Fe(CN{)}_6$

• A. $K_{2}S{O}_4$
• B. $A{l}_2(S{O}_4{)}_3$
• C. $K_{3}Fe(CN{)}_6$
• D. $Al(N{O}_3{)}_3$

Answer 1

The correct option is B $A{l}_2(S{O}_4{)}_3$

Considering all are 100% ionised,
$K_4\left[Fe\right(CN{)}_6]$ dissociates as:
$K_4\left[Fe\right(CN{)}_6]\stackrel{}{⟶}4K^{+}+1[Fe(CN{)}_6{]}^{-4}$
Van't Hoff factor:
$i=5$

$A{l}_2(S{O}_4{)}_3\stackrel{}{⟶}2A{l}^{3+}+3S{O}_4^{-2}$
Van't Hoff factor :
$i=5$

$Al(N{O}_3{)}_3\stackrel{}{⟶}1A{l}^{3+}+3N{O}_3^{-}$
Van't Hoff factor :
$i=4$

$K_{2}S{O}_4\stackrel{}{⟶}2K^{+}+1S{O}_4^{2-}$
Van't Hoff factor :
$i=3$

$K_3\left[Fe\right(CN{)}_6]\stackrel{}{⟶}3K^{+}+1[Fe(CN{)}_6{]}^{-4}$
Van't Hoff factor:
$i=4$

Hence, option (a) is correct.