13 % of solution of sulphuric acid contains 13 g of H2SO4in 100 g of the solution. Weight of water = 100 - 13 = 87 g.
mole of H2SO4 = 13/98 mole
Volume of H2SO4 solution + Weight of soluton / Density = 100/1.02 ml
= 100 / 1.02 x 1/1000 = 1/1.02 litre
Molarity of H2SO4 solution = mole of H2SO4 / Volume of solution
= 13/98 / 1/1.02 = 10.2 x 13/98 = 1.353 M
Molality of H2SO4 solution = mole of H2SO4 / wt of water in kg. = 13/98 / 87/1000 = 13/98 x 1000/87
= 1.525 mol kg-1 (mole Kg-1)
Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as:
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
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.