Can someone please explain this to me?!
Please! Thank you!
How much (in kilograms) ethylene glycol antifreeze (C2H6O2)(molar mass 62.02 g/mol) has to be added to 14.4 kg of water to get a mixture that freezes at -15.5 degrees C. For water Kf = 1.86 degrees C kg/mol
To solve this problem, we need to use the concept of molality and the formula for freezing point depression. Here's how to calculate the amount of ethylene glycol antifreeze required:
1. Determine the molality of the solution:
Molality (m) is defined as the number of moles of solute per kilogram of solvent. In this case, the solvent is water.
We need to find the moles of ethylene glycol required to depress the freezing point. The formula for calculating molality is:
Molality (m) = Moles of solute / Mass of solvent (in kg)
Given that the molar mass of ethylene glycol (C2H6O2) is 62.02 g/mol, we first need to convert the mass of water to moles:
Moles of water (H2O) = Mass of water (in kg) / Molar mass of water
The molar mass of water (H2O) is approximately 18.02 g/mol.
Moles of water = 14.4 kg / 18.02 g/mol
2. Calculate the molality:
Now that we have the moles of water, we can calculate the molality (m) using the formula mentioned earlier:
Molality (m) = Moles of solute / Mass of solvent (in kg)
Since we want to depress the freezing point with ethylene glycol, the solute in this case is ethylene glycol (C2H6O2). We still need to find the moles of ethylene glycol (C2H6O2) required, which can be calculated using its molar mass (62.02 g/mol):
Moles of ethylene glycol = Molality (m) * Mass of water (in kg)
3. Calculate the change in freezing point:
The change in freezing point (∆Tf) is given by the formula:
∆Tf = Kf * Molality
In this equation, Kf represents the molal freezing point depression constant for water, which is given as 1.86 degrees C kg/mol.
Substitute the values we found into the equation:
∆Tf = 1.86 * Molality
Since the mixture needs to freeze at -15.5 degrees C, the change in freezing point (∆Tf) will be the difference between the desired temperature and the freezing point of water (0 degrees C):
∆Tf = -15.5 - 0 = -15.5 degrees C
4. Solve for Molality:
We can now rearrange the equation to solve for Molality (m):
Molality (m) = ∆Tf / Kf
Substitute the values we found:
Molality (m) = -15.5 / 1.86
5. Convert Molality to Moles of ethylene glycol:
Using the formula for Molality (m) from step 2, we can now calculate the moles of ethylene glycol (C2H6O2):
Moles of ethylene glycol = Molality (m) * Mass of water (in kg)
Substitute the values we found:
Moles of ethylene glycol = Molality * 14.4 kg
6. Convert Moles of ethylene glycol to mass:
Finally, we can calculate the mass of ethylene glycol by multiplying the moles by the molar mass (62.02 g/mol):
Mass of ethylene glycol = Moles of ethylene glycol * Molar mass of ethylene glycol
Substitute the values we found:
Mass of ethylene glycol = Moles of ethylene glycol * 62.02 g/mol
This will give you the amount of ethylene glycol antifreeze (in kilograms) needed to be added to 14.4 kg of water to obtain the desired mixture.
To make it easy to type let's call ethylene glycol just eth.
delta T = Kf*m
You know delta T (15.5) and you know Kf, solve for molality eth.
m = mols/kg solvent
You know m eth and kg solvent, solve for mols eth.
mols = grams/molar mass. You know mols eth and molar mass eth, solve for grams eth.