A molar mass that is either lower or higher than the expected or normal value is called as abnormal molar mass.
In order to account for all such abnormalities, introduced a factor
(i) known as van 't Hoff's factor, which represents the extent of association (or) dissociation of a solute.
Van't Hoff's factor is defined as the ratio of the observed colligative property to the calculated colligative property.
i = Observed colligative property / Calculated colligative property
Observed colligative property ∝ 1/Molar Mass
i = Mc/Mo
Van't Hoff's factor
(i) represents the extent of association (or) dissociation of a solute
i = Total number of moles of particles after association or dissociation / Number of moles of particles before association or disscussion
Experimentally determined molar mass is always lower than actual value for dissociation.
Molar Mass ∝ 1/Colligative Property
If the solute undergoes association in a solution, then the value of van 't Hoff's factor is less than one. If the solute undergoes dissociation then 'i' is greater than one.
KCl → K+ + Cl-
1 molecule of KCl furnishes 2 ions in solution
i = Total number of moles of particles after dissociation / Number of moles of particles before dissociation
i = 2/1 = 2
Ethanoic acid Dimer of Ethanoic acid
i = Total number of moles of particles after association / Number of moles of particles before association
i = 1/2 = 0.5
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.
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