Exercise

1. A solution composed of 8.3 g of a non-volatile substance dissolved in 1 mole of chloroform, CHCl3, has a vapour pressure of 511 mmHg. The vapour pressure of purechloroform at this temperature is 526 mmHg.

Calculate:a) the mole fraction of the solute,b) the number of moles of the solute,c) the molar mass of the solute.

Draw a vapour pressure/composition diagram

Benzene C6H6 and toluene C7H8 form nearly ideal solution. At 100 °C, vapour pressures of benzene and toluene are 1390 mmHg and 550 mmHg, respectively. Draw a vapour pressure/composition curves for the benzene – toluene mixture at 100 °C and estimate the composition of the mixture that boils exactly at 100 °C at atmospheric pressure.

Draw a vapour boiling point/composition diagram

The boiling point of benzene is 80 ○C whereas that of ethanolis 78.5 ○C. The vapour pressure when these two componentsare mixed is greater than the vapour pressure of pure benzeneand pure ethanol. An azeotropic mixture with a 32.4 % ethanolcomposition is obtained.

(a) Sketch and label the boiling point-composition phase diagram for the mixture.

(b) Explain the shape of the curve with respect to the attractive forces between molecules.

(c) Name the distillation product if a fractional distillation is carried out on the mixture with a 50 % ethanol composition.

a) What are the boiling points of the pure components?b) What is the more volatile component of the mixture?c) What is the boiling point of the mixture containing equimolar quantities of both components?d) What is the composition of vapour produced by boiling this mixture?e) What is the boiling point of the mixture containing 0.1 molar fraction of n-hexane?f) How many simple distillation steps are required to give distillate containing minimum 0.85 molar fraction of n-hexane?

Colligative Properties

Calculate the boiling point of an aqueous solution containing 138 g of glycerol, C3H5(OH)3, dissolved in 300 g of water. Glycerol is a non-volatile substance and does not dissociate in water.

Calculate the molar mass and molecular formula of a certain chemical compound which does not dissociate in water and which has an empirical formula of C2H3N, knowing that the dissolution of 24.6 g of this substance in 100 g of benzene causes a boiling-point elevation of 5.06 °C.

Calculate the freezing point of a solution prepared by adding 55 g of glucose, C6H12O6, 55 g of sucrose, C12H22O11, to 350 g of water. Kf is equal to 1.861.86 °C/mol.

What is the freezing point of the solution prepared by dissolving 15 g of sucrose,C12H22O11 in 225 g of water, Kf for H2O is equal to 1.86 °C/m ?

A solution of a certain substance in benzene boils at 81.5 °C. What is the freezing point of this solution?

Tb (benzene) = 80.1 °C, Tf (benzene) = 5.5 °C, Kb (benzene) = 2.53 °C/mKf (benzene) = 5.12 °C/m.

Determine the molar mass and formula of an unknown substance.

Solution:16.8 g of solute250 g of waterTf = -0.774 °C , .Tf = 0.774 °CKf (water) = 1.86 °C/m

Solute:57.2% C4.77% H38.1% OMr = ?

The curve shows an azeotropic mixture with a minimum boilingpoint (maximum vapour pressure).

Conclusion: i. The mixture is volatile (a maximum vapour pressure)ii.Intermolecular attraction for ethanol and benzene is weaker than intermolecular attraction within the pure solvents