Question

Which one of the following electrolytes has the same value of van't Hoff's factor (i) as that of $A{l}_2(S{O}_4{)}_3$ (if all are 100% ionised)?

• A. $Al(N{O}_3{)}_3$
• B. $K_4\left[Fe\right(CN{)}_6]$
• C. $K_{2}S{O}_4$
• D. $K_3\left[Fe\right(CN{)}_6]$

Answer 1

The correct option is B $K_4\left[Fe\right(CN{)}_6]$

Van't Hoff factor $i=\frac{\text{number of solute particles present in solution}}{\text{theoretical number of solute particles due to solution of non electrolyte}}=\frac{n\left(observed\right)}{n\left(theoretical\right)}$.

1 molecule of $A{l}_2(S{O}_4{)}_3$ ionizes in solution to produce 5 ions.

$A{l}_2(S{O}_4{)}_3\to 2A{l}^{3+}+3S{O}_4^{2-}$

Hence, $i=\frac{n\left(observed\right)}{n\left(theoretical\right)}=\frac{5}{1}=5$.

1 molecule of $K_4\left[Fe\right(CN{)}_6]$ ionize in solution to produce 5 ions.

$K_4\left[Fe\right(CN{)}_6]\to 4K^{+}+[Fe(CN{)}_6{]}^{4-}$

Hence, $i=\frac{n\left(observed\right)}{n\left(theoretical\right)}=\frac{5}{1}=5$.

Thus, $K_4\left[Fe\right(CN{)}_6]$ has the same value of van't Hoff's factor (i) as that of $A{l}_2(S{O}_4{)}_3$ (assuming 100% ionization).

Note: 1 molecule each of $Al(N{O}_3{)}_3$, $K_{2}S{O}_4$ and $K_3\left[Fe\right(CN{)}_6]$ ionizes in solution to produce 4, 3 and 4 ions respectively.