An ethylene glycol solution contains 26.2g of ethylene glycol (C2H6O2) in 81.4mL of water. Calculate the freezing point of the solution. (Assume a density of 1.00 g/mL for water.)
how do I set this up?
mols ethylene glycol = grams/molar mass = ?
81.4 mL at density of 1.00 g/mL = 81.4g
Convert 81.4 g to kg.
molality = mols eth gly/kg solvent, then
delta T = Kf*m
You know Kf is 1.86 and you know m from above. Solve for delta T and subtract from 0 to find new freezing point.
For the boiling point, Kb is 0.512, m is same as above, solve for delta T = K*m and add to normal boiling point to find new boiling point.
Post your work if you get stuck.
Identify the percent by volume of ethylene glycol in a solution that freezes at -10.°C.
Well, let's break it down:
Since we have a solution of ethylene glycol in water, we can assume the ethylene glycol particles will lower the freezing point of water. The extent to which the freezing point is lowered depends on the concentration of ethylene glycol in the solution.
To calculate the freezing point of the solution, we can use the equation:
ΔTf = Kf * m
Where:
- ΔTf is the change in freezing point
- Kf is the freezing point depression constant (for water, it is approximately 1.86°C/m)
- m is the molality of the solution (moles solute/kg solvent)
First, we need to find the molality of the solution:
Moles of ethylene glycol = mass / molar mass
Moles of ethylene glycol = 26.2 g / (2 * 12.01 g/mol + 6 * 1.01 g/mol + 2 * 15.999 g/mol)
Moles of ethylene glycol = 0.435 mol
Next, we need to find the mass of the water:
Mass of water = volume of water * density of water
Mass of water = 81.4 mL * 1.00 g/mL = 81.4 g
Now, we calculate the molality:
Molality = moles of solute / kg of solvent
Molality = 0.435 mol / 0.0814 kg
Molality = 5.34 m
Finally, we can calculate the change in freezing point:
ΔTf = Kf * m
ΔTf = 1.86°C/m * 5.34 m
ΔTf = 9.91°C
The freezing point depression is equal to the change in freezing point caused by the ethylene glycol. Therefore, to find the freezing point of the solution, we would subtract the freezing point depression from the freezing point of pure water, which is 0°C.
Freezing point of the solution = 0°C - 9.91°C
Freezing point of the solution = -9.91°C
So, the freezing point of the ethylene glycol solution is approximately -9.91°C. Now, don't expect your solution to turn into ice cream, but it'll definitely be colder than pure water!
To calculate the freezing point of the solution, you can use the equation for freezing point depression. This equation relates the change in freezing point to the molality of the solute. The formula is:
∆T = Kf * m
Where:
∆T is the change in freezing point
Kf is the cryoscopic constant of the solvent
m is the molality of the solute
To set up the calculation, you need to determine the molality of the ethylene glycol solution. Here's how you can do it:
1. Calculate the moles of ethylene glycol (C2H6O2):
Number of moles = mass / molar mass
Mass = 26.2 g
Molar mass of C2H6O2 = (12.01 g/mol * 2) + (1.01 g/mol * 6) + (16.00 g/mol * 2) = 62.07 g/mol
Moles = 26.2 g / 62.07 g/mol
2. Calculate the molality of the solution:
Molality = moles of solute / kg of solvent
The mass of water can be calculated using its density and volume:
Mass of water = density of water * volume of water
Density of water = 1.00 g/mL
Volume of water = 81.4 mL = 81.4 g
Molality = moles / (mass of water / 1000)
= moles / (81.4 g / 1000)
Once you have the molality of the solution, you can plug it into the freezing point depression equation along with the cryoscopic constant of water.
Note: The cryoscopic constant of water is 1.86 ºC/m.
∆T = (1.86 ºC/m) * molality
This will give you the change in freezing point (∆T). To calculate the actual freezing point of the solution, subtract the ∆T from the freezing point of pure water, which is 0 ºC.
Calculate the boiling point of the solution.
whats you do for the next step then