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Question

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What are the boiling & freezing point of a solution that contains 28.89 grams of C6H6 in 550.55 grams of solvent CCl4? Boiling point of CCl4 is 76.8 C. The freezing point is -22.8 c. Molal boiling point constant for CCl4 is 5.02 c/m and molal freexing point constant is 29.8 C/m.

Answer 1

Actually

Moles C6H6 = 28.89 g / 78.1121 g/mol = 0.370

molality = 0.370 mol / 0.55055 Kg = 0.672

delta T = kb x m = 5.02 x 0.672 = 3.37 °C

boiling point = 3.37 + 76.8 = 80.2 °C

delta T = kf x m = 29.8 x 0.672 = 20.0 °C

freezing point = - 22.8 - 20.0 = - 42.8 °C

Answer 2

You'll find out more about this in the "colligative properties" section of your textbook.

When a nonvolatile solute is dissolved in a solvent, it will raise the boiling point and reduce the freezing pointof the solution, otherwise known as the boiling elevation and freezing point depression.  The amount that the BP and FP are elevated or reduced depends on the solvent (the constants) and the molality (concentration) of the solute.  And one other thing, the number of particles the solute splits up into if it ionizes.  With benzene in carbon tet, that won't be an issue.

Molality is the concentration in moles of solute per kilogram of solvent.

m = (28.89 g C6H6 x 1 mol C6H6 / 78.0 g C6H6) / 0.55055 kg = 0.6727 m

DTb = kb x m = 5.02 C/m x  0.6727 m

DTb = 3.38 C

Tb' = Ti + DTb = 76.8 + 3.38 = 80.2 C

--------------------------------------...

DTf = kf x m = 29.8 C/m x 0.6727 m = 20.0 C

Tf ' = Tf - DTf = -22.8 - 20C = -44.8 C

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The other part to this is the van't Hoff factor (i) which comes from the number of particles the compound dissociates into.  You see, the effect on BP or FP is due to the *total number of particles in solution*.  If C12H22O11 dissolves in water, there's only one particle for each original molecule.  But if NaCl dissolves in water, there are two particles for every original "molecule", Na+ and Cl-.  Therefore you would get twice the effect, and the BP would go up more and FP would be lowered more.  The van't Hoff factor (i) is included in these forms of the equation.

DT(b) = ik(b)m

and

DT(f) = ik(f)m

Answer 3

The molar mass of C6H6 = 78.0 g/mol.

28.89 g C6H6 x (1 mol / 78.0 g) = 0.370 mol C6H6

"d" = delta (change)

dT = Kf x molality = Kf x mol C6H6 / kg CCl4

dT = 29.8 C/m x 0.370 mol C6H6 / 0.55055 kg CCl4 = 20.0 C

The freezing point is lowered 20.0 C

FP = -22.8 - 20.0 = -42.8 C

dT = Kb x molality = Kb x mol C6H6 / kg CCl4

dT = 5.02 C/m x 0.370 mol C6H6 / 0.55055 kg = 3.37 C

The boiling point is raised by 3.37 C

BP = 76.8 + 3.4 = 80.2 C

Answer 4

Moles C6H6 = 28.89 g / 78.1121 g/mol = 0.370



molality = 0.370 mol / 0.55055 Kg = 0.672

delta T = kb x m = 5.02 x 0.672 = 3.37 °C

boiling point = 3.37 + 76.8 = 80.2 °C

delta T = kf x m = 29.8 x 0.672 = 20.0 °C

freezing point = - 22.8 - 20.0 = - 42.8 °C