contestada

The boiling point of water, H2O, is 100.000 °C at 1 atmosphere. Kb(water) = 0.512 °C/m In a laboratory experiment, students synthesized a new compound and found that when 13.62 grams of the compound were dissolved in 217.5 grams of water, the solution began to boil at 100.094 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound ?

Respuesta :

Edufirst

Answer:

  • 341.1 g/mol

Explanation:

1) Data:

a) Tb₁ = 100.000°C

b) Kb = 0.512 °C/m

c) mass of solute = 13.62 g

d) mass of solvent = 217.5 g

e) Tb₂ = 100.094°C

f) Solute: nonvolatile and nonelectrolyte

g) MM = ?

2) Chemical principles and formulae:

a) The boiling point elevation of a non-volatile solute is a colligative property, which follows this equation:

  • ΔTb = Kb × m × i

Where:

  • ΔTb is the elevation of the boiling point = Tb₂ - Tb₁,
  • Kb is the molal boiling constant of the solvent,
  • m is the molality of the solution,
  • i is the Van't Hoof constant, and is equal to 1 for non-electrolyte solutes.

b) Molality, m:

  • m = number of moles of solute / kg of solvent

c) Molar mass, MM:

  • MM = mass in grams / number of moles

3) Solution:

i) ΔTb = Tb₂ - Tb₁ = 100.094°C - 100.000°C = 0.094°C

ii) ΔTb = Kb × m ⇒ m = ΔTb / Kb = 0.094°C / (0.512°C/m) = 0.1836 m

iii) m = number of moles of solute / kg of solvent ⇒

    number of moles of solute = m × kg of solvent = 0.1836 m × 0.2175 kg

    number of moles of solute = 0.03993 mol

iv) MM = mass in grams / number of moles = 13.62 g / 0.03993 mol = 341.1 g/mol

Otras preguntas