Question

For given molecules :

$\left(I\right)NaN{O}_3\left(II\right)BaC{l}_2\left(III\right)K_4\left[Fe\right(CN{)}_6\left]\right(IV)C_6{H}_{12}{O}_6$
The increasing order of Van't Hoff factor in aqueous solution is :
(Assume complete dissociation wherever applicable)

• A. $\left(I\right)<\left(II\right)<\left(III\right)<\left(IV\right)$
• B. $\left(II\right)<\left(III\right)<\left(I\right)<\left(IV\right)$
• C. $\left(IV\right)<\left(I\right)<\left(II\right)<\left(III\right)$
• D. $\left(IV\right)<\left(III\right)<\left(II\right)<\left(I\right)$

Answer 1

The correct option is C $\left(IV\right)<\left(I\right)<\left(II\right)<\left(III\right)$

For $\left(I\right)$:
$NaN{O}_3\left(aq\right)⟶N{a}^{+}\left(aq\right)+N{O}_3\left(aq\right)$
$i=2$

For $\left(II\right)$:
$BaC{l}_2\left(aq\right)⟶B{a}^{2+}\left(aq\right)+2C{l}^{-}\left(aq\right)$
$i=3$

For $\left(III\right)$:
$K_4\left[Fe\right(CN{)}_6\left]\right(aq)⟶4K^{+}(aq)+[Fe\left(CN{)}_6{]}^{4-}\right(aq)$
$i=5$

For $\left(IV\right)$:
$C_6{H}_{12}{O}_6$ is a non-electrolyte, so it won't dissociate or associate.
$i=1$
So, correct increasing order for Van't Hoff factor is :
$\left(IV\right)<\left(I\right)<\left(II\right)<\left(III\right)$