First, we need to determine the reaction that happens when standardizing. The reactions would be:
KIO3(aq) + 6 H+
(aq) + 5 I-
(aq) ------> 3 I2(aq) + 3 H2O(l) + K+
(aq)
C6H8O6(aq) + I2(aq) ------> C6H6O6(aq) + 2 I- (aq) + 2 H+
(aq)
5g/L C6H8O6 (0.01 L) ( 1 mol / 176.12 g) ( 1 mol I2 / 1 mol C6H8O6 ) ( 1 mol KIO3 / 3 mol I2 ) = 0.00009 mol KIO3
Molarity = 0.00009 mol KIO3 / 0.03037 L = 0.003 M
Hope this answers the question. Have a nice day.
