the vapour pressure of a liquid increases with increase of temperature. It boils at the temperature at which its vapour pressure is equal to the atmospheric pressure. For example, water boils at 373.15 K (100° C) because at this temperature the vapour pressure of water is 1.013 bar (1 atmosphere)
Let be the boiling point of pure solvent and
Tb be the boiling point of solution. The increase in
the boiling point is known as
elevation of boiling point.
for dilute solutions the elevation of boiling point (ΔTb) is directly proportional to the molal concentration of the solute in a solution. Thus
ΔTb ∝ m
or ΔTb = Kb m
Here m (molality) is the number of moles of solute dissolved in 1 kg of solvent and the constant of proportionality, Kb is called Boiling Point Elevation Constant or Molal Elevation Constant (EbullioscopicConstant). The unit of Kb is K kg mol-1. If w2 gram of solute of molar mass M2 is dissolved in w1 gram of solvent, then molality, m of the solution is given by the expression:
Thus, in order to determine M2, molar mass of the solute, known mass of solute in a known mass of the solvent is taken and ΔTb is
determined experimentally for a known solvent whose Kb value is known.
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