Question

Determine the freezing point of a water solution of fructose (C6H12O6) made by dissolving 92.0 g of fructose in 202 g of water.

Answer 1

Answer:

THE FREEZING POINT OF A WATER SOLUTION OF FRUCTOSE MADE BY DISSOLVING 92 g OF FRUCTOSE IN 202g OF WATER IS -4.70 ◦C

Explanation:

To calculate the freezing point of a water solution of fructose,

1. calculate the molar mass of Fructose

( 12 * 6 + 1*12 + 16*6) =72 + 12 +96 = 180 g/mol

2. calculate the number of moles of fructose in the solution

number of moles = mass / molar mass

n = 92 g / 180 g/mol

n = 0.511 moles.

3. calculate the molarity of the solution

molarity = moles / mass of water in kg

molarity = 0.5111 / 202 g /1000 g

molarity = 0.5111 / 0.202

molarity = 2.529 M

4. calculate the change in the freezing point of pure solvent and solution ΔTf

ΔTf = Kf * molarity of the solute

Kf = 1.86 ◦C/m for water

ΔTf = 1.86 * 2.529

ΔTf = 4.70 C

5. the freezing point is therefore

0.00 ◦C - 4.70 ◦C = -4.70 ◦C

Answer 2

Answer:

-4.7°C

Explanation:

In this question, we want to calculate the freezing point of the water solution of fructose.

We proceed as follows;

Firstly, we find the number of moles of the fructose using molar mass

Number of moles = mass/molar mass

mass is 92g , molar mass of fructose is 180g/mol

Number of moles = 92/180 = 0.51 moles

Next thing is to calculate the molality of the solution = number of moles of fructose/mass of water in kg = 0.51/(202/1000) = 0.51/0.202 = 2.525m

Now, we calculate the freezing point depression;

Δ$T_{f}$ = $K_{f}$ × m

where $K_{f}$ refers to the molal freezing point depression of the solvent = 1.86

Δ$T_{f}$ = 1.86 × 2.525 = 4.7 °C

Since the presence of impurities decrease freezing point and the normal freezing point of water is 0 °C, the freezing point of the solution is 0-4.7 = -4.7 °C