Question

0.1 M aqueous solution of K4[fe(CN)6] will have the same freezing point as 0.1 M aqueous solution of

Answer 1

Hi there
0.1 m aqueous K4[fe(CN)6] solution will have same freezing point as that of 0.1 m Ba3(PO4)2 or any other ionic compound that dissosiates into 5 ions like K4[fe(CN)6],because depression in freezing point only depends on number of particles and not their nature.
Hope it helps./

Answer 2

Answer: The correct answer is $Al_{2} (SO_{4})_{3}$  or any other ionic compound that dissociates into 5 ions.

Explanation:    Δ$T_{b} = i.K_{b}.m$
                    For dilute solution, Molarity(M) = molality(m)

                   Concentration i.e is same for all cases i.e 0.1 M for $K_{4} [Fe(CN_{6})]$ also.

      So, the value will depend on $i$ (van’t hoff factor).

      As it is an aqueous solution, $K_{f}$ will also be same.

$i$ for $K_{4} [Fe(CN_{6})]$ —>
                            $K_{4} [Fe(CN_{6})]$   →  $4K^{+}$ $+$ $[Fe(CN)_{6} ]^{+4}$

                    $i$ = 5 ( as it dissociates into 5 ions)

$AlCl_{3}$          →   $2Al^{+3}$ $+$ $3Cl^{-}$            ( $i$ = 4)

Here, $i$ is not same.

$Al_{2} (SO_{4})_{3}$   →   $2Al^{+3}$ $+$  $3(SO_{4})^{-2}$    ( $i$ = 5)

Here, $i$ is same as $K_{4} [Fe(CN_{6})]$ .

So, The freezing point is same as $K_{4} [Fe(CN_{6})]$.