A solution is made by dissolving 0.526 mol of nonelectrolyte solute in 831 g of benzene. Calculate the freezing point and boiling point of the solution. Constants may be found here.
m = 0.526 mol/kg solvent (0.831 kg benzene)
delta T = Kf*m and subtract from zero to find new f.p.
delta T = Kb*m and add to 100 to find new boiling point.
To calculate the freezing point and boiling point of a solution, we need to use the formula given by the equation:
ΔT = K * m
where ΔT is the change in temperature, K is the cryoscopic or ebullioscopic constant, and m is the molality of the solution.
First, let's calculate the molality (m) of the solution:
Molality (m) = moles of solute / mass of solvent (in kg)
Given that 0.526 mol of solute is dissolved in 831 g of benzene, we need to convert the mass of benzene to kilograms:
Mass of benzene = 831 g = 831/1000 kg = 0.831 kg
Now we can calculate the molality:
m = 0.526 mol / 0.831 kg = 0.632 mol/kg
Next, we need to find the cryoscopic constant (Kf) and ebullioscopic constant (Kb) for benzene from the provided constants.
From the constants, we find that the cryoscopic constant (Kf) for benzene is 5.12°C/m and the ebullioscopic constant (Kb) is 2.53°C/m.
Now we can calculate the change in temperature (ΔT) for both the freezing point and boiling point:
For the freezing point (ΔTf):
ΔTf = Kf * m
ΔTf = 5.12°C/m * 0.632 mol/kg = 3.23°C
For the boiling point (ΔTb):
ΔTb = Kb * m
ΔTb = 2.53°C/m * 0.632 mol/kg = 1.60°C
To get the freezing point and boiling point of the solution, we need to add/subtract the calculated values of ΔT to/from the normal freezing point (0°C) and boiling point (100°C) of pure benzene, respectively.
Freezing point of the solution = 0°C - ΔTf = 0°C - 3.23°C = -3.23°C
Boiling point of the solution = 100°C + ΔTb = 100°C + 1.60°C = 101.60°C
Therefore, the freezing point of the solution is -3.23°C and the boiling point of the solution is 101.60°C.