What is the freezing point of a solution that contains 10.0 g of glucose (C6H12O6)in 100.g of H20? Kf for water is 1.86C/m.
What is the freezing point of a solution that contains 10.0 g of glucose
(C6H12O6) in 100 g of H2O?
What is the freezing point of a solution that contains 10.0 g of glucose
(C6H12O6) in 100 g of H2O?
To find the freezing point of the solution, we need to use the formula:
ΔTf = Kf * m
Where:
- ΔTf is the freezing point depression (the difference between the freezing point of the pure solvent and the freezing point of the solution)
- Kf is the molal freezing-point depression constant (given as 1.86 °C/m for water)
- m is the molality of the solution (mol solute/kg solvent)
Step 1: Calculate the molality (m) of the solution.
Molality (m) is defined as the number of moles of solute per kilogram of solvent.
First, we need to calculate the number of moles of glucose (C6H12O6):
Molar mass of glucose (C6H12O6) = 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol
Number of moles = mass / molar mass
Number of moles of glucose (C6H12O6) = 10.0 g / 180.18 g/mol
Next, we need to calculate the mass of water (H2O):
Mass of water = 100.0 g
Now, we can calculate the molality (m) of the solution:
m = moles of solute / mass of solvent (in kg)
m = (10.0 g / 180.18 g/mol) / (100.0 g / 1000) = (10.0 g / 180.18 g/mol) / 0.1 kg
Step 2: Calculate the freezing point depression (ΔTf) by using the formula:
ΔTf = Kf * m
ΔTf = 1.86 °C/m * (10.0 g / 180.18 g/mol) / 0.1 kg
Step 3: Calculate the freezing point of the solution.
The freezing point of the solution can be found by subtracting the ΔTf from the freezing point of the pure solvent.
Freezing point of the solution = Freezing point of pure solvent - ΔTf
The freezing point of pure water is 0 °C.
Freezing point of the solution = 0 °C - ΔTf
Substitute the calculated ΔTf value to find the freezing point of the solution.
moles glucose = grams/molar mass.
molality = moles/kg solvent.
delta T = Kf*molality.