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 42.5 g of barbital in 825 g of acetic acid? K_b = 3.07°C/m; boiling point of pure acetic acid = 117.9°C

Question

Quizplus · Accepted Answer

Firstly, we need to calculate the molality of the solution:

moles of barbital = 42.5 g / 184.2 g/mol = 0.23 mol
mass of acetic acid = 825 g
molar mass of acetic acid = 60.05 g/mol
moles of acetic acid = 825 g / 60.05 g/mol = 13.74 mol
molality = moles of solute / mass of solvent in kg
= 0.23 mol / 0.825 kg
= 0.28 m

Next, we can calculate the boiling point elevation using the formula:

ΔTb = i * Kb * molality

i = van't Hoff factor (assumed to be 1 for molecular solutes)
Kb = boiling point elevation constant for acetic acid = 3.07°C/m

ΔTb = 1 * 3.07°C/m * 0.28 m
ΔTb = 0.86°C

Finally, we can calculate the boiling point of the solution:

boiling point of solution = boiling point of pure solvent + ΔTb
= 117.9°C + 0.86°C
= 118.76°C

Rounding off to the nearest tenth, we get the answer as 118.8°C, which is option D.

Barbiturates Are Synthetic Drugs Used as Sedatives and Hypnotics