Mixing acetone and chloroform occurs with reduction in volume. Name the type of deviation from Raoult's law shown by the above mixture and state whether the process is endothermic or exothermic.

The solution has negative deviation from Raoult's law because decrease in volume indicates strong intermolecular forces of interaction between the molecules of acetone and chloroform. This negative deviation results in reduction of volume of solution as well as the tendency of the molecule to escape from the liquid due to which vapour pressure of the solution decreases. This decrease in the escaping tendency results in increase in the boiling point of the solution. In case of negative deviation, contraction in the volume on mixing along with evolution of heat is expected. Hence, this process is an exothermic reaction.

Calculate the molar mass of a non-volatile solute 1.8 grams of which have been dissolved in 25 grams of acetone and the solution boils at 56.86 C. Pure acetone boils at 56.38 C. The molal elevation constant for acetone is 1.72 k kg per mole

Boiling point of pure acetone is T ' b = 56 . 38 ° C Boiling point of acetone when organic compound is added is T b = 56 . 88 ° C ∆ T b = T b - T ' b = 56 . 86 ° C - 56 . 38 ° C = 0 . 48 ° C Weight of organic compound is W b = 1 . 8 g Weight of acetone is W a = 25 g Molal elevation constant is defined as the elevation in boiling point produced when one mole of solute is dissolved in one kg of the solvent. Molal elevation constant is given as K b = 1 . 72 k kg mol - 1 Molar mass of organic compound is calculated by using the formula, M b = K b × W b × 1000 W a × ∆ T b = 1 . 72 × 1 . 8 × 1000 25 × 0 . 48 = 258 g mole - 1

Draw a graph depicting the deviation from Raoult's law exhibited by a solution prepared by mixing ethanol and acetone.

On mixing equal volumes of acetone and ethanol, positive deviation from Raoult's law is expected. The new interactions are weaker than those in the pure components. The observed vapour pressure of each component and the total vapour pressure are greater than predicted by Raoult's law and hence it shows positive deviation. The graph to depict the deviation is:

The vapour pressure of a 5 % aqueous solution, by mass, of a non-volatile organic substance at 373 K is 745 mm of Hg . Calculate the molar mass of the solute. (Given vapour pressure of water at 373 K is 760 mm Hg .)

5 % aqueous solution of a non-volatile organic substance by mass means, 100 g of solution contains 5 g solute and 95 g solvent (water). Thus, Mass of solute = W b = 5 g Mass of solvent = W a = 95 g Vapour pressure of water at 373 K = p a ° = 760 mm of Hg . Vapour pressure of given solution is = p a = 745 mm of Hg . Molecular mass of solvent (water) = M a = 18 g Molecular mass of solute = M b In a solution the relative lowering of vapour pressure of the solvent is equal to the mole fraction of the non-volatile solute. That is, p a 0 - p a p a 0 = Χ b p a 0 - p a p a 0 = W b × M a W a × M b On substituting the values we get, 760 - 745 760 = 5 . 0 × 18 95 × M B M B = 5 . 0 × 18 × 760 15 × 95 M B = 48 gmol - 1

Draw a neat labelled diagram to show the elevation in boiling point of a solvent when a non-volatile solute is added to it.

When a non-volatile solute is added to a volatile solvent the vapour pressure of the solvent is lowered. The vapour pressure of the solvent in solution is less than 1 atm at the boiling point of pure solvent ( T b ) . Under this condition the liquid will not boil. Because of the addition of nonvolatile solute, the escaping tendency of the solvent molecules from liquid to vapour is diminished as shown by small arrow ( → ) . The liquid-vapour equilibrium is disturbed, and more vapours start condensing back. To restore the equilibrium the vapour pressure should be increased to 1 atm and back condensation should be prevented. Thus, it can be achieved by increasing the temperature from T b to a higher value so that the liquid boils. Thus, the addition of a nonvolatile solute to a volatile solvent elevates it's boiling point.

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