When a solution does not obey Raoult’s law over the entire range of concentration, then it is called non-ideal solution.
The vapour pressure of such solution either be higher or lower. i.e positive when higher
negtive when lower.
The cause for these deviations lie in the nature of interactions at the molecular level.
In case of positive deviation from Raoult’s law,
A-B interactions are weaker than those between A-A or B-B, i.e., in this case the intermolecular attractive forces between the solute-solvent molecules are weaker than those between the solute-solute and solvent-solvent molecules. This means that in such solutions, molecules of A (or B) will find it easier to escape than in pure state. This will increase the vapour pressure and result in positive deviation. Mixtures of ethanol and acetone behave in this manner.
In case of negative deviations from Raoult’s law, the intermolecular attractive forces between A-A and B-B are weaker than those between
A-B and leads to decrease in vapour pressure resulting in negative deviations. An example of this type is a mixture of phenol and aniline.
In this case the intermolecular hydrogen bonding between phenolic proton and lone pair on nitrogen atom of aniline is stronger than the
respective intermolecular hydrogen bonding between similar molecules.
Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Let the total mass of the solution be 100g and mass of benzene be 30 g
therefore mass of tetrachloride= (100-30)g = 70g
Molar mass of benzene,
Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as:
Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.
Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.
(a) Mol. mass of
Volume of solution = 4.3 L
(b) Number of moles present in 1000 ml of 0.5M H2SO4= 0.5 mol
therefore number of moles present in 30ml of 0.5M H2SO4=mol =0.015mol
therefore molarity =0.015/0.5L
thus molarity is 0.03M