Question

The van't Hoff factor is 3 for:

• A. $N{a}_{2}S{O}_4$ with 50% dissociation
• B. $A{l}_2(S{O}_4{)}_3$
• C. $CaC{l}_2$ with 80% dissociation
• D. $K_4\left[Fe\right(CN{)}_6]$ with 50 % dissociation

Answer 1

The correct option is D $K_4\left[Fe\right(CN{)}_6]$ with 50 % dissociation

(A) $N{a}_{2}S{O}_4\rightleftharpoons 2N{a}^{+}+S{O}_4^{2-}$

$1-0.5$ $2\times 0.5$ $0.5$ ($50$% dissociation)

$\therefore$ Vant Hoff factor, $i=1-0.5+2\times 0.5+0.5=2$

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

$i=5$

(C) $CaC{l}_2\rightleftharpoons C{a}^{2+}+2C{l}^{-}$

$1-0.8$ $0.8$ $2\times 0.8$ ($80$% dissociation)

$i=1-0.8+0.8+2\times 0.8=2.6$

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

$1-0.5$ $4\times 0.5$ $0.5$ ($50$% dissociation)

$\therefore i=1-0.5+4\times 0.5+0.5=3$ .