The relative lowering of vapour pressure is given by the following expression.
Where is the vapour pressure of pure solvent, Psolution is the vapour pressure of solution containing dissolved solute, n1 is the number of moles of solvent and n2 is the number of moles of solute.
For dilute solution n2<<n1, therefore the above expression reduces to
Where w1 and w2 are the masses and M1 and M2 are the molar masses of solvent and solute respectively.
We are given that
W1 =30 g
psolution =2.8 kpa
p0solvent =? And M2=?
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:
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,