Determine the freezing point of a 0.51 molal solution of benzoic acid in chloroform. Chloroform has a freezing point of [tex]$-63.5^{\circ} C$[/tex] and a freezing point depression constant of [tex]$4.70^{\circ} C \cdot kg / mol$[/tex].

Formula: [tex]\Delta T_f = K_f m[/tex]

What is the freezing point of this solution? Round to the nearest tenth.

[tex]\boxed{ \quad }^{\circ} C



Answer :

To determine the freezing point of a 0.51 molal solution of benzoic acid in chloroform, we need to use the concept of freezing point depression. The formula we use is:

[tex]\[ \Delta T_f = K_f \cdot m \][/tex]

where:
- [tex]\(\Delta T_f\)[/tex] is the change in the freezing point,
- [tex]\(K_f\)[/tex] is the freezing point depression constant of the solvent,
- [tex]\(m\)[/tex] is the molality of the solution.

Given values are:
- Molality ([tex]\(m\)[/tex]) = 0.51 mol/kg
- Freezing point depression constant ([tex]\(K_f\)[/tex]) = 4.70 [tex]\(^{\circ}C \cdot kg / mol\)[/tex]
- Freezing point of pure chloroform = [tex]\(-63.5^{\circ}C\)[/tex]

First, we need to calculate the change in the freezing point [tex]\(\Delta T_f\)[/tex]:

[tex]\[ \Delta T_f = K_f \cdot m \][/tex]

Substitute the given values into the formula:

[tex]\[ \Delta T_f = 4.70 \cdot 0.51 \][/tex]

The result of this calculation gives us the change in the freezing point:

[tex]\[ \Delta T_f = 2.4^{\circ}C \][/tex]

Next, we determine the freezing point of the solution by subtracting the change in freezing point from the freezing point of the pure chloroform:

[tex]\[ \text{Freezing point of the solution} = \text{Freezing point of pure chloroform} - \Delta T_f \][/tex]

Substitute the known values:

[tex]\[ \text{Freezing point of the solution} = -63.5^{\circ}C - 2.4^{\circ}C \][/tex]

Performing the subtraction gives us:

[tex]\[ \text{Freezing point of the solution} = -65.9^{\circ}C \][/tex]

Therefore, the freezing point of the 0.51 molal solution of benzoic acid in chloroform is [tex]\(-65.9^{\circ}C\)[/tex]. Rounded to the nearest tenth:

[tex]\[ \boxed{-65.9^{\circ}C} \][/tex]