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To make 100.0 mL of a solution that is 0.20 M in potassium iodate using 0.080...

To make 100.0 mL of a solution that is 0.20 M in potassium iodate using 0.080 M sulfuric acid as the solvent requires __________ g of KIO3 (molar mass = 214.001 g/mol).

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Answer #1

molarity of KIO3   = 0.2M

volume of solution = 100ml = 0.1L

no of moles of KIO3 = molarity * volume in L

                                  = 0.2* 0.1 = 0.02moles

mass of KIO3          = no of moles * gram molar mass

                                  = 0.02*214.001    = 4.28002g >>>>answer

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