Answer: The freezing point (in °C) of a 0.743 m aqueous solution of KCI is [tex]2.763^{o}C[/tex].
Explanation:
Given: Molality = 0.743 m
[tex]K_{f}[/tex] for water = [tex]1.86^{o}C/m[/tex]
The equation for dissociation of KCl when dissolved in water is as follows.
[tex]KCl \rightarrow K^{+} + Cl^{-}[/tex]
As it is giving 2 ions. Therefore, Van't Hoff factor for it is equal to 2.
Formula used to calculate the freezing point is as follows.
[tex]\Delta T_{f} = i \times K_{f} \times m[/tex]
where,
i = Van't Hoff factor
[tex]K_{f}[/tex] = molal depression constant
m = molality
Substitute the values into above formula as follows.
[tex]\Delta T_{f} = i \times K_{f} \times m\\= 2 \times 1.86^{o}C/m \times 0.743 m\\= 2.763^{o}C[/tex]
Thus, we can conclude that the freezing point (in °C) of a 0.743 m aqueous solution of KCI is [tex]2.763^{o}C[/tex].
