How many grams of ethylene glycol must be added to 37.8g of water to give a freezing point of -0.150 degrees Celsius
delta T = 0.150
delta T = Kf*m
Substitute dT and Kf and solve for m.
Then m = mols/kg solvent
Substitute ad solve for mols
Then mol = grams/molar mass
Substitute and solve for grams.
Answere Me My Question?
To find the number of grams of ethylene glycol needed to achieve a certain freezing point, we can use the formula for the molal freezing point depression constant (Kf) and the equation for calculating the freezing point depression:
ΔT = Kf * m
Where:
ΔT = change in freezing point
Kf = molal freezing point depression constant
m = molality of the solute
First, let's calculate the molality (m) of the solution.
Molality (m) is defined as the moles of solute per kilogram of solvent. In this case, the solute is ethylene glycol and the solvent is water.
To calculate the molality (m), we need to determine the moles of ethylene glycol and the mass of the solvent (water in this case).
The molar mass of ethylene glycol is 62.07 g/mol.
Moles of ethylene glycol = mass / molar mass
= 37.8 g / 62.07 g/mol
= 0.610 mol
Now, we need to convert the mass of water to kilograms. The density of water is 1 g/mL, so:
Mass of water = 37.8 g
= 37.8 g * (1 kg / 1000 g)
= 0.0378 kg
Now, we need to calculate the change in freezing point (ΔT).
Given: ΔT = -0.150 °C
Finally, we can rearrange the equation and solve for Kf.
Kf = ΔT / m
= -0.150 °C / (0.610 mol / 0.0378 kg)
= -3.365 °C/kg/mol
The molal freezing point depression constant for water is approximately -1.86 °C/kg/mol.
Now, we can calculate the molality of the solution by rearranging the equation to solve for m:
m = ΔT / Kf
= -0.150 °C / -1.86 °C/kg/mol
≈ 0.0806 mol/kg
We now know the molality of the solution. To determine the mass of ethylene glycol needed, we can multiply the molality by the mass of the solvent.
Mass of ethylene glycol = molality * mass of solvent
= 0.0806 mol/kg * 0.0378 kg
≈ 0.00305 kg
Finally, we can convert the mass of ethylene glycol from kilograms to grams:
Mass of ethylene glycol = 0.00305 kg * (1000 g/kg)
≈ 3.05 g
Therefore, approximately 3.05 grams of ethylene glycol must be added to 37.8 grams of water to give a freezing point of -0.150 °C.
To answer this question, we need to use the formula for calculating the freezing point depression:
∆T = Kf * m
where:
∆T is the change in the freezing point,
Kf is the cryoscopic constant (for water, it is 1.86 °C/m), and
m is the molality of the solute.
First, let's calculate the ∆T (change in freezing point):
∆T = -0.150 °C - 0 °C = -0.150 °C
Next, we rearrange the formula to solve for m:
m = ∆T / Kf
Plugging in the values, we have:
m = -0.150 °C / 1.86 °C/m
Now let's calculate the molality:
m = -0.150 °C / 1.86 °C/m ≈ -0.0806 m
Since molality is defined as moles of solute per kilogram of solvent, we need to convert grams to kilograms for both water and ethylene glycol.
Considering 37.8g of water, we convert it to kilograms:
37.8g * (1 kg / 1000g) = 0.0378 kg
To determine the amount of ethylene glycol needed, we need to find the amount in moles using molality and the molecular weight of ethylene glycol (C2H6O2), which is 62.07 g/mol.
First, let's calculate the moles of ethylene glycol:
moles = molality * kg of solvent
moles = -0.0806 m * 0.0378 kg = -0.00305 moles
Now let's convert moles to grams of ethylene glycol:
grams = moles * molecular weight
grams = -0.00305 moles * 62.07 g/mol
Hence, the amount of ethylene glycol that must be added is approximately -0.188 grams. However, negative values do not make sense in this context, so it suggests an error in the calculations or in the initial data. Double-check the calculations or review the problem setup to ensure accuracy.
Note: It's important to exercise caution when working with negative values, as it might indicate an incorrect calculation or an error in the data provided.