What mass of ethylene glycol C2H6O2, the main component of antifreeze, must be added to 10.0 L water to produce a solution for use in a car's radiator that freezes at -23.3C?
Assume the density for water is exactly 1g/mL.
m = moles solute / kg solvent
Help please?
Silly meee!! This question is already answered. :)
I must have worked this problem six times (or more) in the last three days. Three today. Good luck. If you're reading this one note that might help. Instead of going through all the steps that I outlined, there is a short cut that is very useful.
The original way is
delta T = Kf*m
solve for m.
m=mols/kg.
solve for mols.
mols = g/molar mass.
solve for grams.
Here is the shortcut.
delta T = Kf*m
kgsolvent*m*molar mass = grams.
Hope this helps.
neat, huh?
To find the mass of ethylene glycol needed to produce the desired solution, we need to first determine the number of moles of ethylene glycol required. Then, we can convert the moles to mass using the formula m = moles × molar mass.
The molar mass of ethylene glycol (C2H6O2) can be calculated by summing the individual atomic masses of carbon (C), hydrogen (H), and oxygen (O) in the compound:
- Carbon (C) has a molar mass of approximately 12.01 g/mol.
- Hydrogen (H) has a molar mass of approximately 1.01 g/mol.
- Oxygen (O) has a molar mass of approximately 16.00 g/mol.
So the molar mass of ethylene glycol (C2H6O2) can be calculated as follows:
Molar mass of C2H6O2 = (2 × molar mass of C) + (6 × molar mass of H) + (2 × molar mass of O)
= (2 × 12.01 g/mol) + (6 × 1.01 g/mol) + (2 × 16.00 g/mol)
= 24.02 g/mol + 6.06 g/mol + 32.00 g/mol
= 62.08 g/mol
Now, let's find the number of moles of ethylene glycol required using the given formula:
moles solute = (m × kg solvent) / molar mass solute
Since we want the solution to freeze at -23.3°C, we need to convert this temperature to Kelvin (K) using:
Kelvin = Celsius + 273.15
= -23.3°C + 273.15
= 249.85 K
The freezing point depression equation is:
ΔTf = Kf × m
Where:
- ΔTf is the freezing point depression (in K)
- Kf is the cryoscopic constant for water (1.86)
- m is the molality of the solution (moles of solute / kg solvent)
Since the solution freezes at -23.3°C (249.85 K), we can calculate ΔTf as follows:
-23.3 K = 1.86 (mol/kg) × m
Now, rearrange the equation to solve for m:
m = -23.3 K / (1.86 (mol/kg))
= -12.527 m
Since the moles of solute (ethylene glycol) is the same as the molality (m), we can now find the moles of ethylene glycol required:
moles solute = -12.527 mol
Finally, we can find the mass of ethylene glycol using the formula:
mass = moles × molar mass
= -12.527 mol × 62.08 g/mol
= -777.88 g
However, since mass cannot be negative, we can conclude that an ethylene glycol mass of approximately 777.88 g should be added to 10.0 L of water to produce a solution that freezes at -23.3°C.
Please note: The negative sign indicates that the mass should be added, but the direction of the freezing point depression is in the opposite direction.