For given molecules :

$(I) N a N O_{3} (I I) B a C l_{2} (I I I) K_{4} [F e (C N)_{6}] (I V) C_{6} H_{12} O_{6}$
The increasing order of Van't Hoff factor in aqueous solution is :
(Assume complete dissociation wherever applicable)

(I) < (I I) < (I I I) < (I V)

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(I I) < (I I I) < (I) < (I V)

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(I V) < (I) < (I I) < (I I I)

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(I V) < (I I I) < (I I) < (I)

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Solution

The correct option is C $(I V) < (I) < (I I) < (I I I)$
For $(I)$ :
$N a N O_{3} (a q) ⟶ N a^{+} (a q) + N O_{3} (a q)$
$i = 2$

For $(I I)$ :
$B a C l_{2} (a q) ⟶ B a^{2 +} (a q) + 2 C l^{-} (a q)$
$i = 3$

For $(I I I)$ :
$K_{4} [F e (C N)_{6}] (a q) ⟶ 4 K^{+} (a q) + [F e (C N)_{6}]^{4 -} (a q)$
$i = 5$

For $(I V)$ :
$C_{6} H_{12} O_{6}$ is a non-electrolyte, so it won't dissociate or associate.
$i = 1$
So, correct increasing order for Van't Hoff factor is :
$(I V) < (I) < (I I) < (I I I)$

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