Question

Which of the following solutions has the lowest freezing point temperature?
A.$0.1{\rm M}$ $MgC{l_2}$
B.$0.1{\rm M}$ $HCl{O_4}$
C.$0.1{\rm M}$ $N{H_4}OH$
D.$0.1{\rm M}$ $KOH$
E.$0.1{\rm M}$ $LiN{O_3}$

Answer 1

The molarity of each solution given in the options is the same. So freezing point depression will depend upon the number of ions in which each compound dissociates as freezing point is a colligative property.

Complete step by step answer:
Let us first understand depression at a freezing point. Depression in freezing point or elevation in melting point depends upon the number of solute particles rather than the kind of solute particles. These properties of solution are known as colligative properties.
Depression in freezing point $(\Delta {T_f})$ is directly proportional to number of solute particles. Greater the number of solute particles, greater will be the depression in freezing point. It is not necessary that solutions having the same molarity will have the same concentration of solute particles.
So let us see how many ions each compound dissociates upon dissolution. The compound which will dissociate in most numbers of ions will have the lowest freezing point.
First let us see $MgC{l_2}$ . Its dissociation reaction is,
$MgC{l_2} \rightleftharpoons M{g^{2 + }} + 2C{l^ - }$ ,as we can see it dissociates into $3$ ions.
Now let us see $HCl{O_4}$ . Its dissociation reaction is,
$HCl{O_4} \rightleftharpoons {H^ + } + Cl{O_4}^ -$ ,as we can see it dissociates into $2$ ions.
Now let us see $N{H_4}OH$ . Its dissociation reaction is,
$N{H_4}OH \rightleftharpoons N{H_4}^ + + O{H^ - }$ ,as we can see it dissociates into $2$ ions.
Now let us see $KOH$ . Its dissociation reaction is,
$KOH \rightleftharpoons {K^ + } + O{H^ - }$ ,as we can see it dissociates into $2$ ions.
Now let us see $LiN{O_3}$ . Its dissociation reaction is,
$LiN{O_3} \rightleftharpoons L{i^ + } + N{O_3}^ -$ ,as we can see it dissociates into $2$ ions.
Only $MgC{l_2}$ dissociates into $3$ ions, rest all dissociate into $2$ ions. Therefore $MgC{l_2}$ will have the lowest freezing point.

So, option A is the correct answer..

Note:
Apart from boiling point elevation and freezing point depression, osmotic pressure and vapor pressure lowering are also colligative properties. When a solution’s colligative property are different from that calculated theoretically, then the solution is said to have abnormal colligative property.