Consider 100 g of 10% solution of glucose whose conc. is 10% W / W.
Mass of solution = 100 g
Mass of glucose = 10 g
Mass of solvent = 100 – 10 = 90 g
Molar mass of glucose,
C6H12O6 = 6 x 12 + 12 x 1 + 6 x 16
= 72 + 12 + 96 = 180 g / mol
Moles of glucose,
Mole fraction of glucose,
Mole fraction of water
Density of sol = 1.2 g mol L
Therefore, volume of solution
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,
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