When 18 g of ethylene glycol C2H6O2 is dissolved in 150 g of pure water, the freezing point of the solution is
_C . (The freezing point depression constant for water is 1.86C kg)
what i tried was 18/62 to find the moles. then i got .29/.150 to solve for the molality. then i got (.9350)(1.86) to solve for the temperature which is 3.6. i keep getting it wrong im i doing it right? please help. thanks
delta T = Kb*m
m = 18/62 = 0.29 mols
molality = 0.29/0.15 = 1.93 m
delta T = 1.86 x 1.93 = 3.6 degrees.
The normal freezing point is zero, so the new freezing point must be 0.0 - 3.6 = -3.6. I suspect that's your problem. You are stopping at delta T which IS 3.6 but the freezing point is -3.6 C.
thanks sooo much
To find the freezing point of the solution, you need to use the formula for freezing point depression:
ΔT = Kf * m * i,
where ΔT is the change in temperature (freezing point depression), Kf is the freezing point depression constant, m is the molality of the solution, and i is the van't Hoff factor (the number of particles the solute breaks into in the solution).
First, let's calculate the moles of ethylene glycol:
Molar mass of ethylene glycol (C2H6O2) = 2 * atomic mass of carbon + 6 * atomic mass of hydrogen + 2 * atomic mass of oxygen
= (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + (2 * 16.00 g/mol)
= 24.02 + 6.06 + 32.00
= 62.08 g/mol
moles of ethylene glycol = mass of ethylene glycol / molar mass of ethylene glycol
= 18 g / 62.08 g/mol
≈ 0.2901 mol
Next, let's calculate the molality of the solution:
molality (m) = moles of solute / mass of solvent (in kg)
mass of solvent = 150 g = 0.150 kg
molality = 0.2901 mol / 0.150 kg
≈ 1.934 mol/kg
Now, we can calculate the freezing point depression:
ΔT = (1.86 °C kg/mol) * (1.934 mol/kg) * i
The ethylene glycol does not ionize or dissociate in water, so the van't Hoff factor (i) is equal to 1.
ΔT = (1.86 °C kg/mol) * (1.934 mol/kg) * 1
≈ 3.593 °C
Therefore, the freezing point of the solution is decreased by approximately 3.593 °C. To find the freezing point of the solution, subtract this value from the freezing point of pure water (0 °C):
Freezing point of the solution = 0 °C - 3.593 °C
≈ -3.593 °C
So, the freezing point of the solution is approximately -3.593 °C.