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RoseWind [281]
2 years ago
14

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 bo

iling-point-elevation constant for CS2 ?
Chemistry
1 answer:
grandymaker [24]2 years ago
3 0

Answer:

Kb = 0.428 m/°C

Explanation:

To solve this problem we need to use the <em>boiling-point elevation formula</em>:

  • <em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m

Where <em>Tsolution</em> and <em>Tpure solvent</em> 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 <em>calculate the molality of the solution</em>.

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 <u>solve for Kb</u>:

<em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m

  • 47.52 °C - 46.3 °C = Kb * 0.5226 m
  • Kb = 0.428 m/°C
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The equation for the enthalpy change of the given reaction is:

\Delta H_{rxn} =

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