Question

Barbiturates are synthetic drugs used as sedatives and hypnotics. Barbital (184.2 g/mol) is one of the simplest of these drugs. What is the boiling point of a solution prepared by dissolving 50 g of barbital in 625 g of acetic acid?

a. 119.3 b. 117.0 c 118.8 d. 121.2 e. >125C

Answer 1

A solution will boil at a higher temperature than the pure solvent. This is the colligative property called boiling point elevation.

The more solute dissolved, the greater the effect. An equation has been developed for this behavior. It is:

ΔT = i K_bm

The temperature change from the pure to the solution is equal to two constants times the molality of the solution.

The van 't Hoff Factor

The van 't Hoff factor is symbolized by the lower-case letter i. It is a unitless constant directly associated with the degree of dissociation of the solute in the solvent.

Substances which do not ionize in solution, like sugar, have i = 1.

K_b = molal boiling point elevation constant or ebullioscopic constant.

K_b = 3.07°C/m, boiling point of pure acetic acid = 117.9°C

Determine molality of Barbitalin acetic acid solution as follows,

Remember molal is moles solute over kg solvent.

Therefore moles of barbital = 50/184.2 = 0.2714

0.2714 moles of solute / 625 kg of Acetic acid = 4.3431×10^-4m

Molality of solution is 4.3431m

Now, i = 1 (non-ionisable solutes)

Put all values in the equation above ,

Determine bp elevation:

ΔT = i K_bm

     = 1 × 3.07°C/m × 4.3431×10^-4m

     = 0.00133

Boiling point of solution will be= B.P of pure acetic acid + Delta T = 117.9 +0.00133 = 117.901^oC