Question

Of the following 0.10m aqueous solution, which one will exhibit the largest freezing point depression?
A. $KCl$
B. ${{C}_{6}}{{H}_{12}}{{O}_{6}}$
C. $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$
D. ${{K}_{2}}S{{O}_{4}}$

Answer 1

Liquids have a characteristic temperature at which they turn into solids. This temperature is known as freezing point and a temperature at which the vapor pressure is large enough that bubbles form inside the body of the liquid and this temperature is known as boiling point. If pressure on the surface of water increases, its boiling point increases and freezing point decreases.

Complete answer:
Freezing point is a temperature at which liquid starts to convert itself in a solid state i.e. the intermolecular forces gets stronger as we know that intermolecular forces are lesser in liquids due to which they have tendency to flow but by decreasing the temperature forces some close to each other and convert in solid state.
Depression in freezing point can be calculated by the following formula:
$\Delta {{T}_{f}}=i{{K}_{f}}m$
Where ${{K}_{f}}$is freezing point depression constant of solvent and m = molarity
Molarity can be defined as the number of solutes present in 1 liter of the solvent.
I will be equal to Von’t hoff factor which can be calculated by
$\dfrac{Observed\ freezing\ \text{point}}{Calculated\ freezing\ \text{point}}$.
Out of all the options given Van't hoff factor is greater for $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$ which is 5, therefore this implies that it will also have highest depression in freezing point.

Thus option C is the correct answer.

Note:
Depression of freezing point which can be defined as the changes in temperature between solvent and solution ad this can be shown by the formula
$\Delta {{T}_{f}}={{T}^{0}}-{{T}^{s}}$
Where $\Delta {{T}_{f}}$ is depressed in freezing point.
${{T}_{0}}$ = Freezing point of solvent
${{T}_{s}}$= Freezing point of solution