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

Answer 1

118.75°C is the boiling point of a solution.

Step-by-step explanation:

Mass of the solute that is barbiturates = 42.5 g

Molar mass of a solute = 184.2 g/mol

Moles of solute =
`(42.5 g)/(184.2 g/mol)=0.2307 mol`

Mass of the solvent that acetic acid = 825 g = 0.825 kg

`molality=\frac{\text{Moles of solute}}{\text{Mass of solvent}}`

Molality of the solution (m):

`m=(0.2307 mol)/(0.825 kg)=0.2796 m`

Elevation in boiling point is given as:

`\Delta T_b=i* K_b* m`

i = 1 (organic compound)

`=1* 3.07^oC/m* 0.2796 m=0.8585^oC`

`\Delta T_b=T_b-T`

`T_b` = Boiling temperature of solution.

T = boiling temperature of solvent that is acetic acid=117.9°C

`0.8585^oC=T_b-117.9^oC`

`T_b=118.75 ^oC`

118.75°C is the boiling point of a solution.