Question

The solution which has the lowest freezing point temperature is:
a.) $0.1 M Mg{Cl}_{2}$
b.) $0.1 M HCl{O}_{4}$
c.) $0.1 M N{H}_{4}OH$
d.) $0.1 M KOH$
e.) $0.1 M LiN{O}_{3}$

Answer 1

Hint: The freezing point of a temperature depends upon the depression in the freezing point. It also depends upon the concentration of the particles of the solution.

Complete step by step solution:
Freezing point can be defined as the temperature at which a liquid becomes a solid at normal atmospheric temperature. The freezing point depends upon the concentration of particles in a solution. The greater will be the concentration of particles, the lower will be the freezing point. Since, in this case the concentrations of each solution is equal i.e., 1 M, we cannot use this criteria to segregate them. But, we also know that the freezing point of the solution also depends upon the depression in freezing point $(\delta T)$. The depression in freezing point $(\delta T)$ is defined as the decrease in the freezing point on the addition of a non-volatile solute. The greater depression in freezing point $(\delta T)$, the lower will be the corresponding freezing point temperature. It is a colligative property that depends upon the number of solute particles. The greater the number of solute particles , the greater is the depression in freezing point and lower will be the freezing point temperature. Now, let us look at the no. of solute particles of the solutions given.

$0.1 M Mg{Cl}_{2}$: $Mg{Cl}_{2}$ dissociates to give 1 ion of ${Mg}^{2+}$ and 2 ions of ${Cl}^{-}$.
$Mg{ Cl }_{ 2 }\quad \longrightarrow \quad { Mg }^{ +2 }\quad +{ 2Cl }^{ - }$
$0.1 M HCl{O}_{4}$: $HCl{O}_{4}$ dissociates to give 1 ion of ${H}^{+}$ and 1 ion of $Cl{O}_{4}^{-}$.
$HCl{ O }_{ 4 }\quad \longrightarrow \quad { H }^{ + }\quad +\quad { Cl{ O }_{ 4 } }^{ - }$
$0.1 M N{H}_{4}OH$: $0.1 M N{H}_{4}OH$ dissociates to give 1 mole of $N{H}_{3}$ and 1 mole of ${H}_{2}O$.
$N{ H }_{ 4 }OH\quad \longrightarrow \quad N{ H }_{ 3 }\quad +\quad { H }_{ 2 }O$
$0.1 M KOH$: $0.1 M KOH$ dissociates to give 1 ion of ${K}^{+}$ and 1 ion of $O{H}^{-}$.
$KOH\quad \longrightarrow \quad { K }^{ + }\quad +\quad { OH }^{ - }$
$0.1 M LiN{O}_{3}$: $0.1 M LiN{O}_{3}$ dissociates to give 1 ion of ${Li}^{+}$ and 1 ion of $N{O}_{3}^{-}$.
$LiN{ O }_{ 3 }\quad \longrightarrow \quad { Li }^{ + }\quad +\quad { N{ O }_{ 3 } }^{ - }$

Now, we can see that only $Mg{Cl}_{2}$ dissociates to give 3 ions and the rest dissociates to give only 2 ions. Thus, from the relation, we can say it has the highest no. of solute particles which accounts for highest depression in freezing point and thus has the lowest freezing point temperature.

Hence, option (a) is the correct answer.

Note: If the concentration would have been different then we would have used the fact that, 'greater the concentration, the lower will be the freezing point.' Increase in the number of solute particles results in depression of the freezing point while on the other hand the same thing also results in elevation of the boiling point.