± Boiling Point Elevation and Freezing Point Depression for Organic Solutions Th

Question

± Boiling Point Elevation and Freezing Point Depression for Organic Solutions

The temperature at which a solution freezes and boils depends on the freezing and boiling points of the pure solvent as well as on the molal concentration of particles (molecules and ions) in the solution. For nonvolatile solutes, the boiling point of the solution is higher than that of the pure solvent and the freezing point is lower. The change in the boiling for a solution, Tb, can be calculated as

Tb=Kbm

in which m is the molality of the solution and Kb is the molal boiling-point-elevation constant for the solvent. The freezing-point depression, Tf, can be calculated in a similar manner:

Tf=Kfm

in which m is the molality of the solution and Kf is the molal freezing-point-depression constant for the solvent.

Part A

Cyclohexane has a freezing point of 6.50 C and a Kf of 20.0 C/m. What is the freezing point of a solution made by dissolving 0.463 g of biphenyl (C12H10) in 25.0 g of cyclohexane?

Express the temperature numerically in degrees Celsius.

Part B

Paradichlorobenzene, C6H4Cl2, is a component of mothballs. A solution of 2.00 g in 22.5 g of cyclohexane boils at 82.39 C. The boiling point of pure cyclohexane is 80.70 C. Calculate Kb for cyclohexane.

Express the constant numerically in degrees Celsius per molal.

Explanation / Answer

molality = mol of C12H10 / kg of cyclohexane

kg cyclohexaen = 25 g= 25*10^-3 kg

mol of C12H10 = mass/MW= 0.463/154.21 = 0.003

so..

molality = 0.003/( 25*10^-3 ) = 0.12 molal

then

dTf = -K*m = -20*0.12 = -2.4°C

Tf new = 6.5-2.4 = 4.1 °C new

B)

dTb = Kb*m

molal = mol / kg

mol of C6H4Cl2 = 2/147.01 = 0.01360 mol

mass in kg = 22.5 = 22.5*10^-3

molal = 0.01360 /(22.5*10^-3) = 0.604444 molal

so..

dTb = Kb*m

(82.39-80.70) = Kb* 0.604444

Kb = (82.39-80.70) /0.604444

Kb = 2.7959 °C/molal

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