Mass of solute in 300 g of solution of 25% conc.
Mass of solvent in 300 g of solution
= 300 – 75 = 225 g Moles of solute in 400 g of solution of 40% conc.
Mass of solvent in 400 g of solution
= 400 – 160 = 240 g Total mass of the solute
= 75 + 160 = 235 g Total mass of solvent
= 225 g + 240 g = 465 g Total mass of solution
= 300 + 400 = 700 g Therefore, composition of solute in solution after mixing =
Percentage composition of solvent in solution after mixing
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.
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 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,