Question

Weigh by difference the amount needed to prepare 250 mL of a 0.01 M KIO3 solution and quantitatively transfer to a volumetric flask with the aid of a glass funnel. Record its mass to the nearest ±0.1 mg.

Answer 1

Ans. Given,

            Molarity of solution = 0.01 M

            Volume of solution = 250.0 mL = 0.250 L

Required moles of KIO₃ = Molarity x Volume of solution in liters

                                                = 0.01 M x 0.250 L                                       ; [1 M = 1 mol/ L]

                                                = (0.01 mol/ L) x 0.250 L

                                                = 0.0025 mol

Required mass of KIO₃ = Required moles x Molar mass

                                                = 0.0025 mol x (214.00097 g/ mol)

                                                = 0.535002425 g

                                                = 535.0 ± 0.0 mg

# Preparation: Accurately weigh 535.0 ± 0.0 mg KIO₃ in a weighing dish. Transfer the solid into a class A standard volumetric flask 250.0 mL using a clean funnel. Rinse the weighing dish at least three times in the funnel. Rinse the funnel and its tip, too. Make the final volume upto the mark with distilled water. Mix well for uniform concentration. It is the desired solution, 250.0 mL of 0.01 M KIO₃.