Urea, (NH2)2CO, is dissolved in 73.5 g of water. The solution freezes at -0.095°C. How many grams of urea were dissolved to make this solution? (Answer must be in grams)
delta T = Kf*molality
Solve for molality.
molality = moles/kg solvent.
Solve for moles.
moles = grams/molar mass.
Solve for grams.
I'm very grateful to this work and pray that God continue to guide you.It is helping as a chemistry student from the university of Liberia
To solve this problem, we can use the concept of freezing point depression. The freezing point depression is the difference between the freezing point of the pure solvent and the freezing point of the solution. In this case, we have the freezing point of the pure solvent (water) and the freezing point of the solution, so we can calculate the freezing point depression.
First, let's calculate the freezing point depression (∆Tf). The freezing point of pure water is 0°C (273.15 K), and in this case, the solution freezes at -0.095°C. So, the freezing point depression (∆Tf) can be calculated as:
∆Tf = (freezing point of pure solvent) - (freezing point of the solution)
∆Tf = 0°C - (-0.095°C)
∆Tf = 0.095°C
Now, we can use the formula for freezing point depression to solve for the molality (m) of the solution.
∆Tf = Kf * m
In this formula:
- ∆Tf is the freezing point depression
- Kf is the cryoscopic constant (a characteristic property of the solvent, which is water in this case)
- m is the molality (moles of solute per kilogram of solvent)
We need to find the molality (m), so we rearrange the formula:
m = ∆Tf / Kf
The cryoscopic constant for water is 1.86°C/m. Let's plug in the values:
m = 0.095°C / 1.86°C/m
m ≈ 0.051 mol/kg
Now, we can use the molality (m) to find the moles of urea dissolved in water. Since we know the mass of water (73.5 g), we can calculate the moles of urea (n) using the formula:
n = m * (mass of water in kg)
First, we need to convert the mass of water to kilograms:
mass of water (kg) = 73.5 g / 1000
mass of water (kg) = 0.0735 kg
Now, let's calculate the moles of urea:
n = 0.051 mol/kg * 0.0735 kg
n ≈ 0.00374 mol
Finally, to determine the grams of urea dissolved in the solution, we can use the molar mass of urea, which is 60.06 g/mol:
grams of urea = n * molar mass of urea
grams of urea = 0.00374 mol * 60.06 g/mol
grams of urea ≈ 0.224 g
Therefore, approximately 0.224 grams of urea were dissolved in the solution to make it freeze at -0.095°C.