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Vika [28.1K]
3 years ago
8

A sample of glucose ( C6H12O6 ) of mass 8.44 grams is dissolved in 2.11 kg water. What is the freezing point of this solution? T

he freezing point depression constant, Kf , for water is 1.86 °C/mol. (Round your answer to the nearest thousandth)
Chemistry
1 answer:
taurus [48]3 years ago
6 0

Answer:

<em>- 0.0413°C ≅ - 0.041°C (nearest thousands).</em>

Explanation:

  • Adding solute to water causes the depression of the freezing point.

  • We have the relation:

<em>ΔTf = Kf.m,</em>

Where,

ΔTf is the change in the freezing point.

Kf is the freezing point depression constant (Kf = 1.86 °C/m).

m is the molality of the solution.

<em>Molality is the no. of moles of solute per kg of the solution.</em>

  • <em>no. of moles of solute (glucose) = mass/molar mass</em> = (8.44 g)/(180.156 g/mol) = <em>0.04685 mol.</em>

<em>∴ molality (m) = no. of moles of solute/kg of solvent</em> = (0.04685 mol)/(2.11 kg) = <em>0.0222 m.</em>

∴ ΔTf = Kf.m = (1.86 °C/m)(0.0222 m) = 0.0413°C.

<em>∴ The freezing point of the solution = the freezing point of water - ΔTf </em>= 0.0°C - 0.0413°C = <em>- 0.0413°C ≅ - 0.041°C (nearest thousands).</em>

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