what mass of ethylene glycol C2H6O2 the main component of anitfreeze, must be added to 10.0L water to produce a solution for use in a car's radiator that freezes at -23.3°C. Assume the density for water is exactly 1 g/m.L
delta T = Kf*m
You know delta T and Kf (1.86 for water), calculate molality.
molality = mols/kg.
You know molality and kg, solve for moles.
mols = grams/molar mass
You know mols and molar mass, solve for grams. Post your work if you get stuck.
I feel like I have made a mistake because the mass seems too small. I don't know where I went wrong :S
Molality = ÄTf / (Kf * i)
= 23.3°C / (1.86°C/m * 3)
= 4.18m (mol C2H6O2 / kg water)
The amount to dissolve is:
n_C2H6O2 = molality * m_water
= 4.18 mol/kg * 0.1kg
= 0.0418 mol
m_C2H6O2 = n_C2H6O2 * M_C2H6O2
= 0.0418mol * 62g/mol
= 2.59g
Let's see. Why are you using i. i is the van't Hoff factor which is 1 for ethylene glycol.
delta T = Kf*m
23.3 = 1.86*m
m=23.3/1.86 = 12.527 molal.
m = mols/kg solvent
12.527 = mols/10 kg = 125.27 moles.
grams = mols x molar mass = 125.27 x 62.068 = 7,775.18 g ethylene glycol. You need to redo the problem and straighten the significant figures out. I usually carry one more than I'm allowed, then round at the end. I think 23.3 is the most s.f. and 1.86 both have 3 places, so the answer would round to 7.78 x 10^3 grams . Check my work.
I did it again this time I got 7774.51g which is much better. Thanks alot!
I don't understand why is it 10kg solvent?
To solve this problem, we need to use the concept of freezing point depression and determine the molality of the ethylene glycol solution.
The freezing point depression (∆Tf) is given by the equation:
∆Tf = Kf * molality,
where Kf is the freezing point depression constant and depends on the solvent (water in this case) and molality is the amount of solute (ethylene glycol) per kilogram of the solvent.
First, let's calculate the molality:
1. Calculate the amount of water in kilograms:
Since the density of water is given as 1 g/mL, and we have 10.0 L of water, we know that the mass of water is 10.0 L * 1 g/mL = 10.0 kg.
2. Calculate the molality (m):
The molality (m) is defined as moles of solute (ethylene glycol) per kilogram of solvent (water).
We need to convert the amount of ethylene glycol to moles using its molar mass.
The molar mass of C2H6O2 (ethylene glycol) is:
2 * atomic mass of carbon (C) + 6 * atomic mass of hydrogen (H) + 2 * atomic mass of oxygen (O)
= 2 * 12.011 g/mol + 6 * 1.008 g/mol + 2 * 15.999 g/mol
= 62.068 g/mol
Now, we can calculate the molality (m):
m = moles of ethylene glycol / kilograms of water
Since we want the solution to freeze at -23.3°C, we can calculate the change in temperature (∆Tf) as the difference between the freezing point of pure water (0°C) and -23.3°C:
∆Tf = -23.3°C - 0°C = -23.3°C
We also need to find the freezing point depression constant (Kf) for water. For water, Kf = 1.86°C/m.
Now, we can rearrange the formula to solve for moles of ethylene glycol:
moles of ethylene glycol = (∆Tf / Kf) * kilograms of water
Substituting the given values:
moles of ethylene glycol = (-23.3°C / 1.86°C/m) * 10.0 kg
Finally, we can calculate the mass of ethylene glycol using the moles and molar mass:
mass of ethylene glycol = moles of ethylene glycol * molar mass
Substituting the given values, you can find the mass of ethylene glycol required to prepare the solution for use in the car's radiator.