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

Answer:

118.75°C is the boiling point of a solution.

Explanation:

Mass of the solute that is barbiturates = 42.5 g

Molar mass of a solute = 184.2 g/mol

Moles of solute = $\frac{42.5 g}{184.2 g/mol}=0.2307 mol$

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):

$m=\frac{0.2307 mol}{0.825 kg}=0.2796 m$

Elevation in boiling point is given as:

$\Delta T_b=i\times K_b\times m$

i = 1 (organic compound)

$=1\times 3.07^oC/m\times 0.2796 m=0.8585^oC$

$\Delta T_b=T_b-T$

$T_b$ = 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.