Respuesta :
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:
-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[tex]T_{f}[/tex] = [tex]K_{f}[/tex] × m
where [tex]K_{f}[/tex] refers to the molal freezing point depression of the solvent = 1.86
Δ[tex]T_{f}[/tex] = 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
