(i) The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. The ideal solutions have two other important properties. The enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is also zero, i.e.,
ΔmixH = 0
ΔmixV = 0
It means that no heat is absorbed or evolved when the components are mixed. Also, the volume of solution would be equal to the sum of
volumes of the two components.
At molecular level, ideal behaviour of
the solutions can be explained by considering two components A and B. In pure components, the intermolecular attractive interactions will
be of types A-A and B-B, whereas in the binary solutions in addition to these two interactions, A-B type of interactions will also be present.
If the intermolecular attractive forces between the A-A and B-B are nearly equal to those between A-B, this leads to the formation of ideal
example: Solution of n-hexane and n-heptane, bromoethane and chloroethane, benzene and toluene, etc.
Escaping tendency of 'A' and 'B' should be same in pure liquids and in the solution.
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.
Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as:
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,
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