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IgorC [24]
3 years ago
9

A solution is made by dissolving 10.20 grams of glucose (C6H12O6) in 355 grams of water. What is the freezing point depression o

f the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.
Chemistry
1 answer:
Ludmilka [50]3 years ago
3 0

Answer:

0.297 °C

Step-by-step explanation:

The formula for the <em>freezing point depression </em>ΔT_f is

ΔT_f = iK_f·b

i is the van’t Hoff factor: the number of moles of particles you get from a solute.

For glucose,

       glucose(s) ⟶ glucose(aq)

1 mole glucose ⟶ 1 mol particles     i = 1

Data:

Mass of glucose = 10.20 g

  Mass of water = 355 g

                 ΔT_f = 1.86 °C·kg·mol⁻¹

Calculations:

(a) <em>Moles of glucose </em>

n = 10.20 g × (1 mol/180.16 g)

  = 0.056 62 mol

(b) <em>Kilograms of water </em>

m = 355 g × (1 kg/1000 g)

   = 0.355 kg

(c) <em>Molal concentration </em>

b = moles of solute/kilograms of solvent

  = 0.056 62 mol/0.355 kg

  = 0.1595 mol·kg⁻¹

(d) <em>Freezing point depression </em>

ΔT_f = 1 × 1.86 × 0.1595

        = 0.297 °C

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