Raoult established that the lowering of vapour pressure depends only on the concentration of the solute particles and it is independent of their identity.
establishes a relation between vapour pressure of
the solution, mole fraction and vapour pressure of the solvent, i.e.,
The reduction in the vapour pressure of solvent (Δp1) is given as:
In a solution containing several non-volatile solutes, the lowering of the vapour pressure depends on the sum of the mole fraction of different solutes.
Equation 3 can be written as :
The expression on the left hand side of the equation as mentioned
earlier is called relative lowering of vapour pressure and is equal to
the mole fraction of the solute. The above equation can be written as:
Here n1 and n2 are the number of moles of solvent and solute respectively present in the solution. For dilute solutions n2 < < n1, hence neglecting n2 in the denominator we have
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,
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
Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as: