Question

Data: molality (mol/kg) ATfL00) 3.4 13.6 11.8 Scatter Plot: N-Propanol/T-Butanol Molality (m) Series1 .. . Linear (Seriesi) y = 7.2639x R2 = 0.0268 0.5 1 15 ATf (OC)

*even if the datas are wrong, can pinpoint where it's wrong in the answer (will rate, thanks)*

1a. From the graph, determine the freezing point depression constant ?? for t-butanol?

1b. Interpret the molality versus ∆?? graph. What are the trends that you see?

Answer 1

The freezing point of the solvent in a solution will be less than the freezing point of the pure solvent whether the solution contains a non-volatile solute or a volatile solute. However, for simplicity, only non-volatile solutes will be considered here.

Experimentally, we know that the change in freezing point of the solvent in a solution from that of the pure solvent is directly proportional to the molal concentration of the solute:

T = Kfm

where:

        T is the change in freezing point of the solvent,
        Kb is the molal freezing point depression constant, and
        m is the molal concentration of the solute in the solution.

So what you have plotted is " m vs Delta T"

The slope of this curve will give you 1/Kf

From your fitting I can observe, that 1/Kf = 7.26

So Kf = 0.138 °C/m

But

In your case it seems that after reaching some critical concentration, the Delta Tf tends to decrease. Possibly because of some correlations developing between the solute and solvent.

So what I did in getting you the right Kf value is by taking the first 3 points. Plotting ∆Tf vs m linearly fitting it. The slope will get you Kf

Upon doing that I get Kf = 8.16 °C/m which I think is correct when compared to literature.