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

## The normal boiling point of benzene is 80.1 °C. The boiling point elevation constant for benzene is $\text{2.53 °C·kg·mol"^"-1}$

##### 2 Answers
Apr 12, 2015

The boiling point of your solution will be ${84.3}^{\circ} \text{C}$.

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

$\Delta {T}_{b} = {k}_{b} \cdot {b}_{B} \cdot i$, where

$\Delta {T}_{b}$ - the difference between the boiling point of the solution and the boiling point of the pure solvent;
${k}_{b}$ - the ebullioscopic constant of the solvent;
${b}_{B}$ - the molality of the solution;
$i$ - the 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

24.6cancel("g") * "1 mole camphor"/(152.23cancel("g")) = "0.1616 moles camphor"

Now plug your data into the equation for boiling point elevation

$\Delta {T}_{b} = {2.53}^{\circ} \text{C"cancel("kg")/(cancel("mol")) * (0.1616cancel("mol"))/(98.5 * 10^(-3)cancel("kg")) = 4.151^@"C}$

${T}_{\text{sol" - T_"solvent" = 4.151 => T_"sol" = 4.151 + T_"solvent}}$

${T}_{\text{sol" = 4.151 + 80.1 = "84.251"^@"C}}$

Rounded to three sig figs, the answer will be

T_"sol" = color(green)(84.3^@"C")

Apr 13, 2015

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 ${K}_{\text{b}}$ in kg/mol, but I think that I can still help you.

Molality = moles of solute/kilograms of solvent.

$\text{Moles of camphor" = "mass"/"molar mass "= (24.6 cancel("g"))/(152.23 cancel("g")"/mol") = "0.1616 mol}$

$\text{Molality" = "0.1616 mol"/"0.0985 kg" = "1.641 mol/kg}$

Next, I used the formula for the change in the temperature of the boiling point , which is ${K}_{\text{b" × m × "no. of pieces solute dissociates into}}$.

This would equal "2.53 °C·"cancel("kg/mol") × 1.641 cancel("mol/kg") × 1 (since organic substances do not dissociate). The answer is 4.15 °C.

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.