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.

Freezing Point = 7.1 °C

Explanation:

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

°C

The formula for the change in temperature is:

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:

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

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

°C

The freezing point for that solution is:

Freezing Point = 11.9 - 4.77 = 7.1 °C

it's filtration, but this is an assumption so my apologies if i got this