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A solution is made by dissolving 19.5 grams of glucose (C6H12O6) in 0.245 kilograms of water. If the molal freezing point constant for water (Kf) is -1.86 °C/m, what is the resulting Δtf of the solution? Show all he steps taken to solve this problem.

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Respuesta :

superman1987

Answer:

- 0.8217°C.

Explanation:

  • We can solve this problem using the relation:

ΔTf = Kf.m,

ΔTf is the depression in the freezing point.

Kf is the molal freezing point constant for water (Kf = -1.86 °C/m).

m is the molality of glucose.

  • We need to calculate the molality:

Molality (m) is the no. of moles of solute in 1.0 kg of solvent.

∴ m = (mass/molar mass) of glucose/(mass of water (kg))

∴ m = (19.5 g/180.156 g/mol)/(0.245 kg) = 0.4418 m.

∴ ΔTf = Kf.m = (-1.86 °C/m)(0.4418 m) = - 0.8217°C.