Lauryl alcohol is obtained from coconut oil and is used to make detergents. A solution of 5.00 g of lauryl alcohol in 100.0 g of benzene freezes at 4.2 oC. What is the approximate molar mass of lauryl alcohol. Refer to the text book for the boiling point of pure benzene. (80 C)
So this is what I have so far
5.5 C-4.2 C = 1.3 C
OK. You've solve so far for delta T
delta T = Kf*molality
Substitute and solve for molality.
m = mols/kg solvent
Sove for mols.
mols = grams/molar mass You hae grams and mols, solve for molar mass.
To find the molar mass of lauryl alcohol, you can use the freezing point depression equation, which relates the change in freezing point to the molality of the solute.
The equation for freezing point depression is:
ΔT = Kf * m
Where:
- ΔT is the change in freezing point (in degrees Celsius)
- Kf is the freezing point depression constant (a property of the solvent)
- m is the molality of the solute (moles of solute per kilogram of solvent)
First, let's calculate the molality (m) of the lauryl alcohol solution.
Molality (m) is defined as moles of solute per kilogram of solvent, so we need to convert the given masses into their respective units.
Mass of lauryl alcohol = 5.00 g
Mass of benzene = 100.0 g
To convert the mass of benzene to kilograms, divide by 1000:
Mass of benzene in kg = 100.0 g / 1000 = 0.1000 kg
Now we can calculate the molality (m) using the following equation:
m = moles of solute / mass of solvent in kg
Since we don't know the molar mass of lauryl alcohol yet, we'll use the variable "M" to represent it:
m = (5.00 g / M) / 0.1000 kg
Simplifying the equation:
m = 50.0 / (M * 0.1000)
Next, we can calculate the change in freezing point (ΔT) using the given information:
ΔT = 4.2°C - (-5.5°C) (note that the book mentions boiling point but it is irrelevant to this problem)
Simplifying:
ΔT = 4.2°C + 5.5°C = 9.7°C
Now, you need to look up the freezing point depression constant (Kf) for benzene in a reference source. The value of Kf for benzene is typically around 5.12°C/m.
Plugging in the values we have:
9.7°C = 5.12°C/m * (50.0 / (M * 0.1000))
Simplifying:
9.7 = 5.12 * (500/M)
Dividing both sides by 5.12
9.7/5.12 = 500/M
Calculating the approximate molar mass of lauryl alcohol, M:
M ≈ 500 * (5.12/9.7)
M ≈ 264.02 g/mol
Therefore, the approximate molar mass of lauryl alcohol is 264.02 g/mol.