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What is the freezing point of a benzene solution that boils at 98.2 °C?

- The normal boiling point of benzene is 80.1 °C.
- The \(K_b\) value is 2.53 °C/m.
- The normal freezing point of benzene is 5.5 °C.
- The \(K_f\) value is 5.12 °C/m.

Answer :

Final answer:

The freezing point of the benzene solution is approximately 3.54 °C below the normal freezing point of benzene.

Explanation:

Freezing Point of a Benzene Solution

The freezing point depression is calculated using the formula:

ΔTF = Kf × m × i

Where ΔTF is the freezing point depression, Kf is the freezing point depression constant, m is the molality of the solution, and i is the van 't Hoff factor.

In this case, we can calculate the freezing point depression by subtracting the boiling point from the normal boiling point:

ΔTF = 80.1 – 98.2 = -18.1 °C

Next, we can calculate the molality of the solution using the formula:

m = δTb / Kb

Where m is the molality, δTb is the boiling point elevation, and Kb is the boiling point elevation constant.

The boiling point elevation can be calculated by subtracting the normal boiling point from the boiling point:

δTb = 98.2 - 80.1 = 18.1 °C

Finally, we can substitute the values into the freezing point depression formula:

-18.1 = 5.12 × m × i

Since the van 't Hoff factor for a nonelectrolyte like benzene is 1, we can simplify the equation:

-18.1 = 5.12 × m

Now we can solve for m:

m = -18.1 / 5.12 ≈ -3.54 °C/m

Therefore, the freezing point of the benzene solution is approximately 3.54 °C below the normal freezing point of benzene.

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