Urea (NH2)2CO is dissolved in 100.0 g of water. The solution freezes at -0.085 degrees C. how many grams of urea were dissolved to make this solution.
delta T = Kf*molality
Plug in 0.085 for delta T, I assume you know Kf for water, calculate molality.
Then molality x kg solvent x molar mass = grams.
You have all but molar mass which can be calculated.
0.085 degrees C = 1.86 degrees C/m
divide 1.86 and m = .046
then the 100.0g water goes into kg right and then where do i go from here
I would carry the 0.046 to at least three places (I know that is one more place than in the 0.085 number BUT I like to carry one more place and round at the end of the work. However, do as your prof advises you to do). So 0.046 = 0.0457 for now.
molality = mol/kg
0.0457 molal = mols urea/0.1 kg solvent.
mols urea = 0.0457 molal x 0.1 molal = 0.00457.
Then mols = grams/molar mass.
You know mols and you know molar mass. Calculate grams.
yes
To find the number of grams of urea dissolved in the solution, first, you need to calculate the molality of the solution using the formula:
molality = ΔT / Kf
Given that the freezing point depression (ΔT) is -0.085 degrees C and the Kf value for water is 1.86 degrees C/m, you can plug these values into the equation:
molality = -0.085 / 1.86
molality ≈ -0.046
Since you have 100.0 g of water, you need to convert this to kg by dividing by 1000:
mass of solvent (kg) = 100.0 g / 1000
mass of solvent (kg) = 0.1 kg
Now, you can calculate the number of moles of urea using the molality and the mass of solvent:
moles of solute = molality x mass of solvent (kg)
moles of solute = -0.046 x 0.1
moles of solute ≈ -0.0046 mol
Since the negative sign indicates that urea acts as a co-solute, you can ignore it in this calculation.
Finally, to find the number of grams of urea, multiply the moles of solute by the molar mass of urea. The molar mass of urea (NH2)2CO is:
2(14.01 g/mol) + 2(1.01 g/mol) + 12.01 g/mol + 16.00 g/mol
= 60.06 g/mol
grams of solute = moles of solute x molar mass
grams of solute ≈ -0.0046 mol x 60.06 g/mol
grams of solute ≈ -0.276 g
Again, the negative sign can be ignored since mass cannot be negative. Therefore, approximately 0.276 grams of urea were dissolved in the solution.