Calculate the amount of benzoic acid ( C6H5COOH) required for preparing 250 ml of 0.15 mol L-1 Solution in methanol.
Molarity = Mass of solute/ Volume of solution Mass of bezoic acid = Molarity × Volume of solution × Molar mass Volume of solution = 250 mL = 250/1000 = 0.25L Molar mass of benzoic acid C6H5COOH = 7× 12+6×1+2×16= 122gmol-1 Mass of benzoic acid= 0.15 MolL-1× 0.25L × 122g mol-1 = 4.575g Ans.
