1) Benzene freezes at 5.45 degrees celsius. The Kf for benzene is -5.07C/m. What would be the freezing point of a 0.210m solution of octane in benzene?
I am not sure how to start this. I know the equation for freezing point is Kfp * m solute.
So plug in the numbers. That's always a good place to start.
But I caution you that the freezing point is NOT Kfp*m. It's DELTA T that is equal to Kfp*m
would it be delta T = -5.07C/m * 0.210m?
Yes, then the freezing point will be normal f.p. - delta T = new f.p.
To find the freezing point of a solution, you can use the equation: ΔT = Kf * m, where ΔT is the change in freezing point, Kf is the freezing point depression constant, and m is the molality of the solute.
In this case, you are given the freezing point of pure benzene (5.45 degrees Celsius) and the value of Kf for benzene (-5.07°C/m). You need to determine the freezing point of a solution containing 0.210 mol of octane in benzene.
To start, calculate the molality (m) of the octane solution. Molality is defined as the number of moles of solute divided by the mass of the solvent (in kilograms). Since both benzene and octane are non-polar, they can be considered a solution.
The molarity of the solution can be calculated using the formula:
m = moles of solute / mass of solvent (in kg)
As we know, moles = mass / molar mass.
Mass of octane = moles of octane * molar mass of octane
Mass of benzene = moles of benzene * molar mass of benzene
Suppose the mass of octane is "m1" and the mass of benzene is "m2".
Since the molecular formula of octane is C8H18, the molar mass of octane (Molar_mass_octane) is:
12 * 8 + 1 * 18 = 114 g/mol = 0.114 kg/mol
Similarly, suppose the molar mass of benzene (Molar_mass_benzene) is x kg/mol.
Using the above information, the equation for moles of octane can be written as:
mol of octane = m1 / Molar_mass_octane ...(1)
Similarly, the equation for moles of benzene is:
mol of benzene = m2 / Molar_mass_benzene ...(2)
Now, since we are given that the solution is 0.210 m, we can write the equation for molality as:
m = (mol of octane) / (moles of benzene + mol of octane) ...(3)
Substituting equations (1) and (2) into equation (3), we get:
0.210 = (m1 / Molar_mass_octane) / (m2 / Molar_mass_benzene + m1 / Molar_mass_octane)
Rearranging the equation, we have:
0.210 = Molar_mass_benzene / Molar_mass_octane * (m1 / (m2 + m1))
Now, we need to solve for m2 (mass of benzene). Rearranging the equation gives:
m2 = (Molar_mass_benzene/molar_mass_octane) * (m1 / (0.210 - m1))
Once you determine the value of m2 (mass of benzene), substitute it into the equation: ΔT = Kf * m.
Finally, calculate the change in freezing point (ΔT) by multiplying Kf (-5.07°C/m) by the molality (m) calculated above. Then, subtract the calculated ΔT from the freezing point of pure benzene (5.45°C) to find the freezing point of the octane in benzene solution.