Which solution has the lower freezing point?
(a)
90.0 g CH3OH in 100. g H2O
180.0 g CH3CH2OH in 200. g H2O
(b)
70.0 g H2O in 1.00 kg CH3OH
70.0 g CH3CH2OH in 1.00 kg CH3OH
delta T = Kf m
convert 90 g CH3OH to mols
mols/kg solvent = m (kg solvent = 0.1)
do the same for 180 g C2H5OH and 200 g H2O.
Follow the same procedure for b.
Why don't you use the Van't Hoff factor for these? I don't get when to use them and when not to.
The Van't Hoff factor can be used, if you know it (them); however, in these cases none of the materials are ionizable. Methanol and ethanol (the first two) are organic compounds that don't ionize at all. In the second set both solutions are in methanol and that doesn't ionize. The van't Hoff factor is useful when NaCl, or Na2SO4 (or something like those two) dissolve in water and form 2 particles (in the case of NaCl) or 3 particles (in the case of Na2SO4.). In that case, I add and i to the equation but it is one for these four problems. Does this clear this up?
Yes, that clears up the use of the Van't Hoff factor. The Van't Hoff factor is used when a solute dissociates into multiple particles in a solution, hence increasing the effective concentration of the solute. In this case, since the solutes are organic compounds that don't ionize, the Van't Hoff factor is not applicable.
To determine which solution has the lower freezing point, we can use the equation delta T = Kf * m, where delta T is the change in freezing point, Kf is the cryoscopic constant, and m is the molality of the solution.
First, let's calculate the molality for each solution:
(a) 90.0 g CH3OH in 100. g H2O:
Molar mass of CH3OH = 32.04 g/mol
Number of moles of CH3OH = 90.0 g / 32.04 g/mol = 2.808 mol
Mass of water = 100. g
Molality (m) = 2.808 mol / 0.100 kg = 28.08 mol/kg
180.0 g CH3CH2OH in 200. g H2O:
Molar mass of CH3CH2OH = 46.07 g/mol
Number of moles of CH3CH2OH = 180.0 g / 46.07 g/mol = 3.909 mol
Mass of water = 200. g
Molality (m) = 3.909 mol / 0.200 kg = 19.54 mol/kg
(b) 70.0 g H2O in 1.00 kg CH3OH:
Molar mass of CH3OH = 32.04 g/mol
Number of moles of CH3OH = 1000 g / 32.04 g/mol = 31.21 mol
Mass of water = 70.0 g
Molality (m) = 70.0 g / 31.21 mol/kg = 2.24 mol/kg
70.0 g CH3CH2OH in 1.00 kg CH3OH:
Molar mass of CH3CH2OH = 46.07 g/mol
Number of moles of CH3CH2OH = 70.0 g / 46.07 g/mol = 1.519 mol
Mass of CH3OH = 1000 g
Molality (m) = 1.519 mol / 1.00 kg = 1.519 mol/kg
Now, we can compare the molality values for each solution. The lower the molality, the lower the freezing point.
In case (a), Solution 1 has a molality of 28.08 mol/kg and Solution 2 has a molality of 19.54 mol/kg. Therefore, Solution 2 (180.0 g CH3CH2OH in 200. g H2O) has the lower freezing point.
In case (b), Solution 3 has a molality of 2.24 mol/kg and Solution 4 has a molality of 1.519 mol/kg. Therefore, Solution 4 (70.0 g CH3CH2OH in 1.00 kg CH3OH) has the lower freezing point.
To summarize:
(a) Solution 2 (180.0 g CH3CH2OH in 200. g H2O) has the lower freezing point.
(b) Solution 4 (70.0 g CH3CH2OH in 1.00 kg CH3OH) has the lower freezing point.