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gayaneshka [121]
3 years ago
13

What is the freezing point of a solution of 0.5 mol of LiBr in 500 mL of water? (Kf = 1.86°C/m) –1.86°C –7.44°C –5.58°C –3.72°C

Chemistry
1 answer:
tresset_1 [31]3 years ago
6 0

Answer:

  • Last choice: <em><u>- 3.72°C</u></em>

Explanation:

The freezing point depression in a solvent is a colligative property: it depends on the number of solute particles.

The equation to predict the freezing point depression in a solvent is:

  • ΔTf = Kf × m × i

Where,

  • ΔTf is the freezing point depression of the solvent,
  • m is the molality,
  • Kf is the cryoscopic molal constant of the solvent, and i is the Van'f Hoff factor, which is the number of ions produced by each unit formula of the ionic compound.

The calcualtions are in the attached pdf file. Please, open it by clicking on the image of the file.

Download pdf
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Answer:

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Explanation:

<u>1) General explanation:</u> a <em>potential chemical energy diagram</em> is used to show how the <em>reactants</em> gain energy until they reach the <em>activation energy</em>, form the <em>activated complex</em>, and release part of the energy to form the <em>products</em>.

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The molar ratio of Br⁻ to Br₂ is 5:3, that is, when 5 moles of Br⁻ are consumed, 3 moles of Br₂ are produced. If the average rate of consumption of Br⁻ is 1.20 × 10⁻⁴ M/s, the average rate of formation of Br₂ is:

\frac{1.20molBr^{-}}{L.s} .\frac{3molBr_{2}}{5molBr^{-}} =\frac{0.720molBr_{2}}{L.s}

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