Question

The depression in freezing point of 0.1M aqueous solutions of ${\text{HCl, CuS}}{{\text{O}}_{\text{4}}}{\text{ and }}{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ are in the ratio:
A.$1:1:1.5$
B.$1:2:3$
C.$1:1:1$
D.$2:4:3$

Answer 1

The decrease in freezing point for a solution is known as its depression in freezing point. The addition of non volatile solute results in the depression of freezing point of a solution. The depression in freezing point depends upon the concentration of solute particles.

Complete step by step answer:
The temperature at which a fluid (liquid or gas) solidifies or freezes is known as its freezing point. The depression in freezing point indicates the decrease in freezing point for a substance, when we add a non volatile solute in any given solvent then the freezing point for that solution decreases. The examples of non volatile solutes (substances that readily do not evaporate at given conditions are called non volatile substances) are like salt in water, alcohol in water etc.
The depression in freezing point is basically a colligative property, because it depends upon the ratio of solute particles to the solvent particles present in the solution. The depression in freezing point depends upon the concentration of solute particles.
In chemistry the formula used to find depression in freezing point is as follows:
${\Delta}{{\text{T}}_{\text{f}}}{\text{ = i}}{{\text{K}}_{\text{f}}}{\text{m}}$
Where, ${\Delta }{{\text{T}}_{\text{f}}}$ is depression in freezing point, $i$ represents the number of ions produced and ${\text{m}}$ is the concentration of solute.
In this problem we have provided that the concentration for all the species is same i.e. 0.1M
Therefore,
${\Delta }{{\text{T}}_{\text{f}}} \propto {\text{i}}$
And we can see that ions produced by ${\text{HCl}}$ will be two, therefore ${\text{i = 2}}$ ,
Ions produced by ${\text{CuS}}{{\text{O}}_{\text{4}}}$ are two, therefore ${\text{i = 2}}$
And ions produced by ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ will be three, therefore ${\text{i = 3}}$
Therefore, the ratio for depression in freezing point will be $2:2:3$
Therefore; ${\text{HCl:CuS}}{{\text{O}}_{\text{4}}}{\text{:}}{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \equiv 1:1:1.5$

Hence option (A) is correct.

Note: the depression in freezing point as we can see in the formula (${\text{\Delta }}{{\text{T}}_{\text{f}}}{\text{ = i}}{{\text{K}}_{\text{f}}}{\text{m}}$) depends mainly upon two factors, ${\text{i}}$ i.e. the number of ions produced by the solute in the solution and ${\text{m}}$, the concentration of solute particles.