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Calculate the freezing point of a solution containing 25 grams of KCl and 2750.0 grams of water. The molal freezing point depression constant (Kf) for water is 1.86 ∘C/m.1.23 oC-0.45 oC+0.45 oC-0.23 oC+0.23 oC

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Answer:

-0.45 °C

Explanation:

The freezing point depression can be calculated using the formula:

  • ΔT=Kf * b * i

Where ΔT is the change of the freezing point temperature (from 0°C of pure water), Kf is 1.86 °C/m, b is the molality, and i is 2 (in this case with KCl).

All data required is given except for b, so we calculate the molality:

  • molality = moles solute / kilograms of solvent

25 g KCl ÷ 74.55g/mol = 0.3353 mol KCl

2750 g water ⇒ 2750/1000 = 2.75 kg water

  • molality = 0.3353/2.75 = 0.1219 m

Now we calculate ΔT:

  • ΔT = 1.86 °C/m * 0.1219 m * 2
  • ΔT = 0.45 °C

So the freezing point of the solution is [0°C - 0.45°C] = -0.45 °C

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