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Colligative Properties-
Ethylene glycol (C2H6O2) is the principal ingredient in antifreeze. How many grams of ethylene glycol will be needed to lower the freezing point of 2100 grams of water by 20 degrees celcius?
delta T = Kf*m
Solve for m = molality
m = mols/kg solvent
You have m and kg solvent; solve for mols.
mols = grams/molar mass
You have mol and molar mass, solve for grams.
Note the correct spelling of celsius.
To solve this problem, you can use the concept of colligative properties and the formulas related to them. Colligative properties are physical properties of a solvent that depend on the number of solute particles, regardless of their identity.
The freezing point depression (ΔTf) is a colligative property, which can be calculated using the formula:
ΔTf = Kf * m * i
Where:
- ΔTf is the change in freezing point
- Kf is the cryoscopic constant, which is a constant specific to a particular solvent (in this case, water)
- m is the molality of the solute
- i is the van 't Hoff factor, which represents the number of particles the solute dissociates into in solution
In this case, ethylene glycol is the solute and water is the solvent. The molar mass of ethylene glycol is 62.07 g/mol, and we need to find the mass to lower the freezing point by 20 degrees Celsius.
First, we need to calculate the molality (m) of the ethylene glycol solution using the given information. Molality is the concentration of the solute in the solvent expressed in moles of solute per kilogram of solvent.
1. Calculate the moles of ethylene glycol:
moles = mass / molar mass
2. Calculate the kilograms of water:
mass of water = 2100 g / 1000 = 2.1 kg
3. Calculate the molality:
m = moles of ethylene glycol / mass of water
After obtaining the molality, you can use the formula for freezing point depression to solve for the mass of ethylene glycol.
ΔTf = Kf * m * i
Rearranging the formula, we can solve for the mass of ethylene glycol:
mass of ethylene glycol = (ΔTf / (Kf * i)) * mass of water
Substitute the values into the formula and compute.
Remember to consult the cryoscopic constant (Kf) for water and the van 't Hoff factor (i) specific to ethylene glycol.
Once you have the value of the mass of ethylene glycol, you will have the answer to the problem.
To calculate the amount of ethylene glycol needed to lower the freezing point of water, we can use the formula for the freezing point depression:
ΔTf = Kf * m
Where:
ΔTf = change in freezing point
Kf = cryoscopic constant (which is specific for each solvent)
m = molality of the solute (moles of solute per kilogram of solvent)
In this case, we are given:
ΔTf = 20 degrees Celsius (since the freezing point of water is 0 degrees Celsius)
Kf for water = 1.86 °C/m
m = ?? (unknown)
First, we need to calculate the molality (m) using the given mass of water and the molar mass of ethylene glycol:
1. Calculate the number of moles of ethylene glycol:
Number of moles = Mass / Molar mass
Molar mass of ethylene glycol (C2H6O2) = 2 * atomic mass of carbon (12.01 g/mol) + 6 * atomic mass of hydrogen (1.01 g/mol) + 2 * atomic mass of oxygen (16.00 g/mol)
2. Calculate the molality (m):
Molality = Moles / Mass of water (converted to kilograms, since molality is defined as moles of solute per kilogram of solvent)
Finally, we can plug in the values into the freezing point depression equation to solve for the mass of ethylene glycol needed:
ΔTf = Kf * m
20°C = 1.86 °C/m * m
Solving for m gives us the molality of the ethylene glycol. Then, we can use it to calculate the mass of ethylene glycol using the equation:
Mass of ethylene glycol = m * Mass of water (converted to kilograms) / Molar mass of ethylene glycol
Let's calculate the values step by step.