Question

Moles of ascorbic acid in 25ml and 500ml in lemonade. (Procedure B)

Very stuck, need help please. ASAP! Thank you

4. Moles of ascorbic acid in 25.00 mL Lemonade 5. Moles of ascorbic acid in 500.00 mL Lemonade.

Answer 1

Volume of KIO₃ needed to react with ascorbic acid: 14.30 mL = 0.0143 L

Molarity of KIO₃ solution = 0.001 M

Moles of KIO₃ = Molarity * Volume of the solution = 0.001 M * 0.0143 L = 1.43 * 10^-5 mole

According to reaction stoichiometry:

Moles of ascorbic acid : Moles of KIO3 = 3:1

Or, Moles of Ascorbic acid = 3 * Moles of KIO3 = 3 * 1.43 * 10^-5 mole = 4.29 * 10^-5 mole

The average volume of lemonede: (60 + 80 + 75 + 75) mL/ 4 = 72.5 mL

So, moles of Ascorbic acid in 25.00 mL of lemonede: 4.29 * 10^-5 mole * ( 25 / 72.5) = 1.48 * 10^-5 mole

moles of Ascorbic acid in 500.00 mL of lemonede: 4.29 * 10^-5 mole * ( 500 / 72.5) = 2.96 * 10^-4 mole