An ethylene glycol solution contains 18.2g of ethylene glycol (C2H6O2) in 85.4mL of water.
Calculate the freezing point and boiling point of the solution. (Assume a density of 1.00 g/mL for water.)
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what's the answer
To find the freezing point and boiling point of a solution, we need to consider the properties of the solute and solvent.
First, let's calculate the molality of the solution, which is the amount of moles of the solute per kilogram of the solvent. The formula for molality (m) is:
m = (moles of solute) / (mass of solvent in kg)
Step 1: Calculate the moles of ethylene glycol (C2H6O2):
To find moles, we will use the formula:
moles = (mass) / (molar mass)
The molar mass of ethylene glycol (C2H6O2) is:
2(12.01 g/mol) + 6(1.01 g/mol) + 2(16.00 g/mol) = 62.07 g/mol
So, moles of ethylene glycol = 18.2 g / 62.07 g/mol = 0.2934 mol
Step 2: Calculate the mass of water in kg:
Since the density of water is 1.00 g/mL, we have 85.4 mL of water. To convert this to kg, we divide by 1000.
mass of water = 85.4 mL * 1.00 g/mL / 1000 = 0.0854 kg
Step 3: Calculate the molality (m):
m = 0.2934 mol / 0.0854 kg ≈ 3.437 mol/kg
Now, we can use the molality to calculate the freezing point and boiling point depressions of the solution.
The formula to calculate the freezing point depression and boiling point elevation is:
∆T = Kf/m for freezing point ∆T = Kb/m for boiling point
Kf and Kb are the cryoscopic and ebullioscopic constants, which depend on the properties of the solvent. For water, the values are:
Kf = 1.86 °C/m
Kb = 0.512 °C/m
Step 4: Calculate the freezing point depression:
∆T(freezing) = Kf * m
∆T(freezing) = 1.86 °C/m * 3.437 mol/kg ≈ 6.39182 °C
The freezing point of pure water is 0 °C, so the freezing point of the solution would be:
Freezing point = 0 °C - 6.39182 °C ≈ -6.39 °C
Step 5: Calculate the boiling point elevation:
∆T(boiling) = Kb * m
∆T(boiling) = 0.512 °C/m * 3.437 mol/kg ≈ 1.761344 °C
The boiling point of pure water is 100 °C, so the boiling point of the solution would be:
Boiling point = 100 °C + 1.761344 °C ≈ 101.76 °C
Therefore, the freezing point of the solution is approximately -6.39 °C and the boiling point is approximately 101.76 °C.
mols ethylene glycol = grams/molar mass = ?
Then m = molality = mols/kg solvent (kg solvent is 0.0854 in the problem)
The formula are similar for f.p. and b.p.
f.p. delta T = Kf*m. Substitute and solve for delta T, then subtract from the normal f.p. of O C to find the new f.p.
b.p. delta T = Kb*m. Substitute and solve for delta T and add to the normal point point of water which is 100 C to find the new boiling point.
Post your work if you get stuck.