Question

What is the freezing point of a 1,4-dioxane solution if that same solution boils at 104.6 °C? The normal boiling point and freezing point of pure 1,4-dioxane (C4H8O2) are 101.5 °C and 11.9 °C, respectively. For 1,4-dioxane, Kb= 3.01 °C/m and Kf= 4.63 °C/m.

Answer 1

Answer:

Freezing Point = 7.1 °C

Explanation:

The boiling point changed from 101.5 °C to 104.6 °C

$\Delta T= 104.6 - 101.5 = 3.1$°C

The formula for the change in temperature is:

$\Delta T=iK_{b}m$

Where:

i: van't Hoff factor

K_{b}: constant

m: molality

In this problem it is impossible to know the van't Hoff factor and the molality alone, but you can determinate how much is their multiplication:

$\ 3.1=3.01im \\im = 1.0299$

Similar to the Boiling Point Elevation, the formula for the change of temperature for the Freezing Point Depression is:

$\Delta T=iK_{f}m$

Replacing im for 1.0299 and K_{f} for 4.63:

$\Delta T=4.63*1.0299$

$\Delta T=4.8$°C

The freezing point for that solution is:

Freezing Point = 11.9 - 4.77 = 7.1 °C