How do you calculate the mass of KIO3 and the volume of 0.100 M H2SO4 needed to prepare 50.0 mL of a .2 M potassium iodate and .08 M sulfuric acid solution?

Answer 1

(1) Mass of KIO3 needed to prepare 50.0 ml of 0.2 M solution:

molar mass KIO3 = 214.0 g/mole

50.0 ml = 0.0500 L and 0.2 M = 0.2 moles/L

mass needed = 0.0500 L x 0.2 moles/L x 214 g/mole = 2.14 g = 2 g (to 1 significant figure)

(2) Volume of 0.100 M H2SO4 needed to prepare 50 ml 0.08 M H2SO4:

V1M1 = V2M2

V1 = (x ml)

M1 = 0.100 M

V2 = 50 ml

M2 = 0.08 M

(x ml)(0.100 M) = (50 ml)(0.08 M)

x = 40. mls needed (to 2 significant figures)