contestada

When 0.270 mol of a nondissociating solute is dissolved in 410.0 mL of CS2, the solution boils at 47.52 ∘C. What is the molal boiling-point-elevation constant for CS2 ?

Respuesta :

maacastrobr

Answer:

Kb = 0.428 m/°C

Explanation:

To solve this problem we need to use the boiling-point elevation formula:

  • Tsolution - Tpure solvent = Kb * m

Where Tsolution and Tpure solvent are the boiling point of the CS₂ solution (47.52 °C) and of pure CS₂ (46.3 °C), respectively. Kb is the constant asked by the problem, and m is the molality of the solution.

So in order to use that equation and solve for Kb, first we calculate the molality of the solution.

molality = mol solute / kg solvent

  • Density of CS₂ = 1.26 g/cm³
  • Mass of 410.0 mL of CS₂ ⇒ 410 cm³ * 1.26 g/cm³ = 516.6 g = 0.5166 kg

molality = 0.270 mol / 0.5166 kg = 0.5226 m

Now we solve for Kb:

Tsolution - Tpure solvent = Kb * m

  • 47.52 °C - 46.3 °C = Kb * 0.5226 m
  • Kb = 0.428 m/°C

Otras preguntas