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# For which of the following solutes the van’t Hoff factor is not greater than one?(A) $NaN{O_3}$(B) $BaC{l_2}$(C) ${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$(D) $N{H_2}CON{H_2}$

Last updated date: 18th Sep 2024
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Hint: As we know that the ionic compounds are polar compounds which can dissolve in water and dissociates into its ions by obeying the law of conservation of mass. The non-polar compounds do not dissociate into water so will not give any ions.

The van’t-Hoff factor is represented as $i$and it is used to modify the calculation of colligative property.
Suppose we have sodium chloride as a solute and we dissolve this into water to know lowering in vapour pressure of solution. When we dissolve the sodium chloride into solvent, it dissociates into ions of sodium chloride, but we were assuming one solute particle as sodium chloride. So, to know the accurate value of lowering in vapour pressure we have to multiply the values by the total number of dissociated ions. Therefore, the number of dissociated ions of solute is van’t Hoff factor ($i$).
Let’s calculate the number of ions of given solutes-
$NaN{O_3}$when dissociates into water it will give two ions as
$NaN{O_3} \to N{a^ + }(aq) + N{O_3}^ - (aq)$
Therefore, the van’t Hoff factor ($i$)$= 2$
$BaC{l_2}$ when dissociates into water it will give three ions as
$BaC{l_2} \to B{a^{2 + }}(aq) + 2C{l^ - }(aq)$
Therefore, the van’t Hoff factor ($i$)$= 3$
${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$when dissociates into water it will give five ions as
${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right] \to 4{K^ + }(aq) + {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }}(aq)$
Therefore, the van’t Hoff factor ($i$)$= 5$
$N{H_2}CON{H_2}$does not dissociate into water because it is a non-polar compound.
Therefore, the van’t Hoff factor ($i$)$= 1$
Therefore, the correct option is option (D).
Note:
The Van't Hoff factor is also defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.