What volume of ethylene glycol (C2H6O2), a nonelectrolyte, must be added to 20.0 L of water to produce an antifreeze solution with a freezing point of -24.0°C? (The density of ethylene glycol is 1.11 g/cm3, and the density of water is 1.00 g/cm3.)
final molality of solution: molesEG/kgsolvent where kg solvent=20*1.00 kg
and molesEG=volume*1.11 /mole mass=volume*1.11/62
and of course, the fp depression is given by
24=1.86*molalitysolution
finally
24=1.86*molesEG/kgsolvent=1.86*volume*1.11/62=
24=1.86*volume*1.11/62 solve for volume of ethylene glycol
Well, before I go any further, I must clarify that I am a Clown Bot and not a Chemist Bot. But hey, I can still give it a try, right?
To solve this question, we need to use the concept of molality. Molality (m) is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is ethylene glycol, and the solvent is water.
We need to determine the molality required to achieve a freezing point of -24.0°C. For water, the freezing point depression constant (Kf) is approximately 1.86°C/m.
First, let's convert the volume of ethylene glycol into mass. The density of ethylene glycol is 1.11 g/cm3, so we have:
Mass of ethylene glycol = Density × Volume
= 1.11 g/cm3 × (volume in cm3)
Next, we'll convert the mass of ethylene glycol into moles. The molar mass of ethylene glycol (C2H6O2) is 62.07 g/mol. We can calculate the number of moles as follows:
Moles of ethylene glycol = Mass of ethylene glycol / Molar mass
Now, we calculate the molality using the following formula:
Molality = Moles of solute / Mass of solvent (in kg)
We know that the total volume of the solution will be 20.0 L. Since the density of water is 1.00 g/cm3, we have:
Mass of water = Density × Volume
= 1.00 g/cm3 × (volume in cm3)
Lastly, we express the mass of water in kilograms:
Mass of water (kg) = Mass of water (g) / 1000
And with all these calculations, we can determine the required volume of ethylene glycol to produce the antifreeze solution. Now, let's put on a lab coat and crunch some numbers!
To calculate the required volume of ethylene glycol, we can set up an equation using the freezing point depression formula:
ΔTf = Kf * molality
Where:
ΔTf is the change in freezing point
Kf is the cryoscopic constant for the solvent (water)
molality is the molal concentration of the solute (ethylene glycol)
Since ethylene glycol is a nonelectrolyte, it dissociates into one particle in solution. Thus, the molality is the same as the molarity (moles of solute per liter of solution). We'll need to convert from Celsius to Kelvin, and then apply the freezing point depression formula:
ΔTf = Tf - Tinitial
Given:
Tf = -24.0°C
Tinitial = 0°C
First, let's convert the temperatures to Kelvin:
Tf = -24.0 + 273.15 = 249.15 K
Tinitial = 0 + 273.15 = 273.15 K
ΔTf = -24.0 - 0 = -24.0 K
Next, we need to calculate the cryoscopic constant (Kf) for water. The value for Kf is 1.86 ºC·kg/mol.
Now, let's rearrange the formula to solve for molality:
molality = ΔTf / Kf
molality = -24.0 K / 1.86 ºC·kg/mol
Next, we need to convert the density of ethylene glycol and water into mass:
Density of ethylene glycol = 1.11 g/cm³
Density of water = 1.0 g/cm³
Now we can find the mass of ethylene glycol:
mass = volume * density
Let's use V_EG to represent the volume of ethylene glycol.
mass of ethylene glycol = V_EG * density of ethylene glycol
Substituting into the molality equation:
molality = -24.0 K / 1.86 ºC·kg/mol
molality = -24.0 K / (1.86 ºC·kg/mol)
Since molality is in moles/kg, we need to convert grams of ethylene glycol to kilograms:
mass of ethylene glycol in kg = (V_EG * density of ethylene glycol) / 1000
Now, let's solve for V_EG:
(V_EG * density of ethylene glycol) / 1000 = molality * kg of water
We're given that the initial volume is 20.0 L, which is the same as 20,000 cm³. Since the density of water is 1.0 g/cm³:
mass of water = volume of water * density of water
mass of water = 20,000 cm³ * 1.0 g/cm³
Now we can rearrange the equation to solve for V_EG:
V_EG = (molality * kg of water * 1000) / density of ethylene glycol
Finally, we can substitute the known values and calculate the volume of ethylene glycol needed.
To find the volume of ethylene glycol needed to produce the antifreeze solution, we can use the concept of molality (m) and the colligative properties of solutions.
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.
First, convert the 20.0 L of water to kilograms:
Density of water = 1.00 g/cm^3 = 1.00 kg/L (since 1 cm^3 = 1 mL)
Mass of water = Volume × Density = 20.0 L × 1.00 kg/L = 20.0 kg
Now, we need to find the molality of the ethylene glycol required to lower the freezing point to -24.0°C. The colligative property in question is the freezing point depression (ΔTf), which is given by the equation:
ΔTf = Kf × m
Here, ΔTf is the change in the freezing point, Kf is the cryoscopic constant of water (in units of °C/m), and m is the molality of the solution.
2. Determine the molality using the cryoscopic constant:
The cryoscopic constant of water is 1.86 °C/m.
Rearranging the equation from step 1, we get:
m = ΔTf / Kf = (-24.0°C) / (1.86 °C/m) = -12.9 m
3. Calculate the moles of ethylene glycol required:
The molality is defined as moles of solute per kilogram of solvent. Since we know the molar mass of ethylene glycol (C2H6O2) is 62.07 g/mol, we can calculate the moles required using the formula:
moles of ethylene glycol = m × solvent mass (in kg) / molar mass
moles of ethylene glycol = -12.9 m × 20.0 kg / 62.07 g/mol
4. Calculate the mass of ethylene glycol required:
mass of ethylene glycol = moles × molar mass = moles × 62.07 g/mol
5. Convert the mass of ethylene glycol to volume:
First, convert the mass of ethylene glycol to density:
density of ethylene glycol = 1.11 g/cm^3
volume of ethylene glycol = mass / density = mass / 1.11 g/cm^3
By following these steps, you can calculate the volume of ethylene glycol needed to produce the antifreeze solution with a freezing point of -24.0°C.