calculate the freezing point of an aqueous solution of 15.5 grams of glucose dissolved in 150 grams of water
To calculate the freezing point of the aqueous solution, we can use the equation:
∆Tf = Kf * m
where ∆Tf is the freezing point depression, Kf is the cryoscopic constant for water (given as 1.86 °C·kg/mol), and m is the molality of the solution.
First, let's calculate the molality (m) of the solution using the formula:
m = moles of solute / mass of solvent in kg
The given mass of glucose is 15.5 grams, and we need to convert it to moles using the molar mass of glucose. The molar mass of glucose (C6H12O6) is approximately 180.16 g/mol.
moles of glucose = mass / molar mass = 15.5 g / 180.16 g/mol
Next, we need to convert the mass of water to kilograms:
mass of water = 150 grams = 150 / 1000 kg
Now, we can calculate the molality:
m = (15.5 g / 180.16 g/mol) / (150 / 1000 kg)
Calculate the value inside the brackets first:
m = 0.08592 mol / 0.15 kg = 0.5732 mol/kg (rounded to four decimal places)
Now that we have the molality (m), we can calculate the freezing point depression (∆Tf):
∆Tf = (1.86 °C·kg/mol) * (0.5732 mol/kg)
∆Tf = 1.07 °C (rounded to two decimal places)
Finally, to find the freezing point of the solution, we subtract the calculated freezing point depression (∆Tf) from the freezing point of pure water, which is 0 °C.
Freezing point of the solution = 0 °C - 1.07 °C
The freezing point of the aqueous solution of 15.5 grams of glucose dissolved in 150 grams of water is approximately -1.07 °C.
To calculate the freezing point of an aqueous solution, we need to use the concept of molality and the freezing point depression constant (Kf) for water, which is equal to -1.86 °C/m.
First, we need to convert the given masses of glucose and water into moles using their molar masses.
The molar mass of glucose (C6H12O6) is 180.16 g/mol.
Moles of glucose = mass of glucose / molar mass of glucose
= 15.5 g / 180.16 g/mol
= 0.086 moles
The molar mass of water (H2O) is 18.015 g/mol.
Moles of water = mass of water / molar mass of water
= 150 g / 18.015 g/mol
= 8.328 moles
Next, we calculate the molality (m) of the solution, which is defined as the moles of solute per kilogram of solvent.
Molality (m) = moles of solute / mass of solvent in kg
= 0.086 moles / (150 g water / 1000)
= 0.573 mol/kg
Now we can use the freezing point depression equation to calculate the change in freezing point (ΔT).
ΔT = Kf * m
= -1.86 °C/m * 0.573 mol/kg
= -1.06 °C
Finally, we can calculate the freezing point of the solution:
Freezing point of water = 0 °C
Freezing point of solution = Freezing point of water - ΔT
= 0 °C - (-1.06 °C)
= 1.06 °C
Therefore, the freezing point of the aqueous glucose solution is 1.06 °C.
mols glucose = grams/molar mass = ?
Then m = mols glucose/kg solvent
Then delta T = Kf*m
New freezing point = zero C - delta T.