Question

1a) How many mg is 5.95×10–5 mol of ascorbic acid (i.e. Vitamin C)? Report your answer to 1 decimal place.

b)What is the concentration of a potassium iodate solution after you complete the following porcedure? Pipette 24 mL of a 0.16 M potassium iodate stock solution to a 100-mL volumetric flask and then add water to the mark. Report your answer in decimal form to 3 decimal places.

Answer 1

Ascorbic Acid [C6H8 067 C=12 H=1, 0 = 16 Atomic mass Molar mass of Acid = 12x6 + 8x1 + 6x16 - 72 + 8 + 96 = 176 g/mol Given, mole = 5.95 xlors mass = 5.95 x 10-5 molar mass mass = 5.95x105 x 176 F = 1047.2x 155 g - 1047 2x105 x 103 mg mass - 10.472 mg)

Holarity = 0.16M, Volume = 24 mL mole of k2O3 = mokesity & Volume ( L) - 0.16 x 24x153 New Volume after water added = 100 ML - 100 X 183 L Molarity = z mole Volume 0.16*24 X 1003 1oox 103 - 10.0384 M New concentration