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

The freezing point of the 1,4-dioxane solution is approximately 7.13 °C, calculated by finding the molality using the boiling point elevation and then applying the freezing point depression constant.

Step-by-step explanation:

To determine the freezing point of a 1,4-dioxane solution using its boiling point elevation and freezing point depression properties, we can use the colligative properties equations. Since the solution boils at 104.6 °C, which is higher than the normal boiling point of pure 1,4-dioxane (101.5 °C), we can calculate the boiling point elevation (ΔTb). Using this elevation and the boiling point elevation constant (Kb), we can find the molal concentration (m) of the solution.

ΔTb = observed boiling point - normal boiling point
ΔTb = 104.6 °C - 101.5 °C = 3.1 °C

Next, we can calculate the molality (m):
m = ΔTb / Kb
m = 3.1 °C / 3.01 °C/m = 1.03 m

Now, we can use the molality and the freezing point depression constant (Kf) to find the freezing point depression (ΔTf):
ΔTf = Kf * m
ΔTf = 4.63 °C/m * 1.03 m = 4.7669 °C

We subtract ΔTf from the normal freezing point to find the freezing point of the solution:
Freezing Temperature = normal freezing point - ΔTf
Freezing Temperature = 11.9 °C - 4.7669 °C ≈ 7.1331 °C

The freezing point of the 1,4-dioxane solution is therefore approximately 7.13 °C.

Answer 2

Freezing Point = 7.1 °C

Step-by-step 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