Question

What is the boiling point of an aqueous solution that freezes at -2.05 degrees C? Kfp = 1.86 K/m and Kbp = 0.512 K/m. Enter your answer using 2 decimal places!!!!

Answer 1

To solve this problem we will apply the concepts of Boiling Point elevation and Freezing Point Depression. The mathematical expression that allows us to find the temperature range in which these phenomena occur is given by

Boiling Point Elevation

$\Delta T_b = K_b m$

Here,

$K_b$ = Constant ( Different for each solvent)

m = Molality

Freezing Point Depression

$\Delta T_f = K_f m$

Here,

$K_f =$ Constant ( Different for each solvent)

m = Molality

According to the statement we have that

$\Delta T_f = T_0 -T_f = 0-(-2.05)$

$\Delta T_f = 2.05\°C$

From the two previous relation we can find the ratio between them, therefore

$\frac{\Delta T_b}{\Delta T_f} = \frac{K_b}{K_f}$

$\frac{\Delta T_b}{\Delta T_f} = \frac{0.512}{1.86}$

$\frac{\Delta T_b}{\Delta T_f} = 0.275$

We already know the change in the freezing point, then

$\Delta T_b = 0.275 (\Delta T_f)$

$\Delta T_b = 0.275 (2.05)$

$\Delta T_b = 0.5643\°C$

The temperature difference in the boiling point is 100°C (Aqueous solution), therefore

$T_b -100 = 0.5643$

$T_b = 100.56\°C$

Therefore the boiling point of an aqueous solution is $100.56\°C$