# What is the boiling point of a solution of 24.6 g of camphor in 98.5 g of benzene?

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The normal boiling point of benzene is 80.1 °C.

The boiling point elevation constant for benzene is #"2.53 °C·kg·mol"^"-1"#

The normal boiling point of benzene is 80.1 °C.

The boiling point elevation constant for benzene is

##### 2 Answers

The boiling point of your solution will be

So, you're dealing with a boiling point elevation problem in which an added solute, in your case camphor, will increase the boiling point of a pure solvent, benzene.

The equation for boiling point elevation is

**van't Hoff factor** - takes into account the number ions a compound forms in solution.

Since camphor is a covalent compound, it will not dissociate when placed in the solvent, which implies a van't Hoff factor equal to **1**.

Since molality is defined as moles of solute per kilogram of solvent, you need to determine how many moles of camphor you have in **24.6 g**

Now plug your data into the equation for boiling point elevation

Rounded to three sig figs, the answer will be

Okay, I will try to answer this but I may be off due to rounding off numbers. I am also a little concerned about the

**Molality = moles of solute/kilograms of solvent.**

Next, I used the formula for the **change in the temperature of the boiling point , which is #K_"b" × m × "no. of pieces solute dissociates into"#.**

This would equal

If you add the change to the original boiling point of 80.1 °C, you get the new boiling point of 84.3 °C.

I am not sure that all of the math is correct, but you have a couple formulas that might help you.