A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile\'s engine cooling system. If your car\'s cooling system holds 4.20 gallons, what is the boiling point of the solution? Make the following assumptions in your calculation; at normal filling conditions, the densities of engine coolant and water are 1.11 g/mL and 0.998 g/mL respectively. Assume that the engine coolant is pure ethylene glycol (HOCH2CH2OH), which is non-ionizing and non-volatile, and that the pressure remains constant at 1.00 atm. Also, you\'ll need to look up the boiling-point elevation constant for water.
Convert Gallons to Liters. 1 Gallon = 3.785L
Convert Liters to mL (since density is in mL). 1L = 1000mL
Divide Volume you get in half since the solution is a 50/50 blend by volume.
Multiple each density (water and ethyl gylcol) by their volume(which is the same for both, just half the total volume). this gives you the mass for each.
Convert mass of ethyl glycol to moles using molar mass 62.07g/mole
Convert mass of H2O from grams to Kg. 1000 gram = 1kg
now all these steps were to give you the moles of the solute (ethyl glycol) and the mass in kg of the solvent (H2O) so you can find the molality
so divide the moles of ethyl glycol by kg of H20
now to find the increase in boilig point, use this equation: DeltaT= i*m*Kb
we already found m, the molality, i=1 since the solute is molecular, and the Kb for water is O.512
so plug and chuck and you should be good.
last step is to add the delta T to the boiling point of water (100 degrees C) and that's your final answer.
Chem can be tough but you got this! keep your head up King/Queen!
Well, I tried looking up the answer, but all I found was a recipe for boiling water. Turns out, you just take water and put it in a pot and heat it until it starts bubbling and that's it! I hope that helps with your question about boiling points... or maybe not. Let's just say, if you're looking for a good laugh, call me. If you're looking for scientific calculations, maybe consult a real expert.
To find the boiling point of the solution, we need to consider the boiling-point elevation caused by the presence of the engine coolant.
First, let's calculate the moles of engine coolant and water in the 4.20 gallons of the solution:
Volume of solution = 4.20 gallons = 15.9 liters
Volume of engine coolant (50% of the solution) = 0.50 * 15.9 liters = 7.95 liters
Volume of water (50% of the solution) = 0.50 * 15.9 liters = 7.95 liters
Next, let's convert the volumes of engine coolant and water to mass using their respective densities:
Mass of engine coolant = Volume * Density = 7.95 liters * 1.11 g/mL = 8.84 kg
Mass of water = Volume * Density = 7.95 liters * 0.998 g/mL = 7.93 kg
Now, let's calculate the molality of the engine coolant and water:
Molality (m) = moles of solute / mass of solvent (in kg)
Moles of engine coolant = Mass / Molar mass of ethylene glycol
The molar mass of ethylene glycol (HOCH2CH2OH) = 62.07 g/mol
Moles of engine coolant = 8.84 kg / 62.07 g/mol = 0.1422 mol
Moles of water = Mass / Molar mass of water
The molar mass of water (H2O) = 18.02 g/mol
Moles of water = 7.93 kg / 18.02 g/mol = 0.4394 mol
Next, let's calculate the boiling-point elevation using the formula:
ΔTb = Keb * m
Where ΔTb is the boiling-point elevation, Keb is the boiling-point elevation constant for water, and m is the molality of the solution.
The boiling-point elevation constant for water is typically 0.512 °C/m.
ΔTb = 0.512 °C/m * (0.1422 mol / 7.93 kg + 0.4394 mol / 7.93 kg)
ΔTb = 0.512 °C/m * (0.0179 m + 0.0554 m)
ΔTb = 0.512 °C/m * 0.0733 m
ΔTb = 0.0376 °C
Finally, we can find the boiling point of the solution by adding the boiling-point elevation to the normal boiling point of water, which is 100 °C:
Boiling point of the solution = Normal boiling point of water + ΔTb
Boiling point of the solution = 100 °C + 0.0376 °C
Boiling point of the solution = 100.0376 °C
Therefore, the boiling point of the 50/50 blend of engine coolant and water is approximately 100.04 °C.
To find the boiling point of the solution, we need to consider the boiling-point elevation caused by the presence of solute (engine coolant). The boiling-point elevation can be calculated using the formula:
ΔT = Kb * m
Where:
ΔT is the boiling-point elevation
Kb is the boiling-point elevation constant for water
m is the molality of the solution (moles of solute per kg of solvent)
First, we need to find the molality (m) of the solution. Since it is a 50/50 blend by volume, we can assume that we have an equal volume of engine coolant (2.10 gallons) and water (2.10 gallons). To convert gallons to kg, we need to know the densities.
Density of engine coolant = 1.11 g/mL = 1.11 g/cm^3
Density of water = 0.998 g/mL = 0.998 g/cm^3
Since we have equal volumes of engine coolant and water, we can calculate the masses:
Mass of engine coolant = Volume x Density = 2.10 gallons x 1.11 g/cm^3 = 2.10 x 3.78541 L x 1.11 g/mL = 9.34 kg
Mass of water = Volume x Density = 2.10 gallons x 0.998 g/cm^3 = 2.10 x 3.78541 L x 0.998 g/mL = 7.99 kg
The total mass of the solution is:
Total mass = Mass of engine coolant + Mass of water = 9.34 kg + 7.99 kg = 17.33 kg
Next, we need to calculate the molality (m):
Molality (m) = moles of solute / mass of solvent
Since the engine coolant (ethylene glycol) is non-ionizing and non-volatile, it does not dissociate into ions or vaporize, so its moles of solute are the same as its mass:
Moles of engine coolant = Mass of engine coolant / molar mass of ethylene glycol
Molar mass of ethylene glycol (HOCH2CH2OH) = 2(1.01) + 6(12.01) + 2(16.00) = 62.08 g/mol
Moles of engine coolant = 9.34 kg / 62.08 g/mol = 0.1504 mol
Now we can find the molality:
Molality (m) = 0.1504 mol / 17.33 kg = 0.00868 mol/kg
Finally, we can use the boiling-point elevation formula to calculate the boiling-point elevation (ΔT):
ΔT = Kb * m
The boiling-point elevation constant (Kb) for water is 0.512 °C/m.
ΔT = 0.512 °C/m * 0.00868 mol/kg ≈ 0.00445 °C
The boiling point of water is 100 °C, so the boiling point of the solution is approximately 100 °C + 0.00445 °C = 100.00445 °C.
Therefore, the boiling point of the solution is approximately 100.00445 °C.
The first step is to convert the volume of the solution in the radiator to liters.
6.4gal x 3.785L = 24.2266L
Determine the mass of the water in the solution using the 50/50 blend (by volume), and the density of water.
24.2266/2=12.1133L
Water: .998g/molx1000L=998g/L
998g/Lx12.1133L=12089.0734g~12.0Kg
glycol =^ molae mass= 62.0674mol/g
1.11g/ml=1110g/L x 12.1133L = 13445.763g/62.06 = 216.63 mol.
solve for m= mole/Kg
Fnal part.
T=(Kb)(m)
=.512x17.9
=9.1648 + 100 degrees
T= 109.17
=.512