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 #"2.53 °C·kg·mol"^"-1"#

2 Answers
Apr 12, 2015

The boiling point of your solution will be #84.3^@"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

#DeltaT_b = k_b * b_B * i#, where

#DeltaT_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

#DeltaT_b = 2.53^@"C"cancel("kg")/(cancel("mol")) * (0.1616cancel("mol"))/(98.5 * 10^(-3)cancel("kg")) = 4.151^@"C"#

#T_"sol" - T_"solvent" = 4.151 => T_"sol" = 4.151 + T_"solvent"#

#T_"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_"b"# in kg/mol, but I think that I can still help you.

Molality = moles of solute/kilograms of solvent.

#"Moles of camphor" = "mass"/"molar mass "= (24.6 cancel("g"))/(152.23 cancel("g")"/mol") = "0.1616 mol"#

#"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_"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.