By using the formula of depression in freezing, it can easily identify the rank of the given compounds.
[tex]\Delta T= i.K_{f}.m[/tex]
where,
[tex]\Delta T[/tex] = freezing point depression
i = van 't Hoff factor (number of ions per individual molecule of solute)
m = molality of the solute
[tex]K_{f}[/tex] = molal freezing point
According to question, if concentration (m) and [tex]K_{f}[/tex] is same, then only van 't Hoff factor will change.
For [tex]NH_{4}Cl[/tex], i = 2
For [tex]Na_{3}PO_{4}[/tex], i = 4
For [tex]Li_{2}SO_{4}[/tex], i= 3
Now, the solute with the largest i value results in low freezing point.
The highest value of i is 4 ([tex]Na_{3}PO_{4}[/tex]), thus [tex]Na_{3}PO_{4}[/tex] has lowest freezing point.
And, [tex]NH_{4}Cl[/tex] has highest freezing point.
The order of freezing point is : [tex]NH_{4}Cl> Li_{2}SO_{4}> Na_{3}PO_{4}[/tex]