Question

A glucose solution contains 60.8 g of glucose (C6H12O6) in 495 g of water. Part A Calculate the freezing point of the solution. (Assume a density of 1.00 g/mL for water.) Express your answer using three significant figures.

Answer 1

The freezing point of the solution is -1.27°C

Step-by-step explanation:

Colligative property of freezing point depression, to solve this.

ΔT = Kf . m

ΔT = T° fussion of pure solvent - T° fussion of solution

Kf = Cryoscopic constant ( For water = 1.86 °C/m)

m = molality (moles of solute in 1kg of solvent)

T° fussion of pure solvent = 0°C

0°C - T° fussion of solution = 1.86°C / m . m

In 495 g of water, we have 60.8 g of glucose

We must apply a rule of three, to find out the grams of glucose in 1000 g of water.

495 g H₂O _____ 60.8 g C₆H₁₂O₆

1000 g H₂O _____ (1000 . 60.8) /495 = 122.82 g C₆H₁₂O₆

Now, let's convert 122.82 in moles through the molar mass.

Mass / Molar mass = Moles

122.82 g / 180.15 g/m = 0.681 m

0°C - T° fussion of solution = 1.86°C / m . 0.681 m

T° fussion of solution = - (1.86°C / m . 0.681 m)

T° fussion of solution = - 1.27°C