Lab 21 Vitamin C Titratio Prelab Exercises for Lab 21 Vitamin C Titration Name S

Question

Lab 21 Vitamin C Titratio Prelab Exercises for Lab 21 Vitamin C Titration Name Section Date 1. If the serving size on the label of juice is 8 oz., and the label claims that the juice has 120% of the RDI per serving, how many mg AA/100 mL is claimed? (29.6 mL 1 fluid oz.) lao%. Vitamin C 1209 per loomL serving AA/toomla 1a0 x 1osmo l.aximlseru 2. If you measure 0.0532 g of AA in Part 1 of the procedure and you titrate 5.22 mL of the l2 solution, what is the concentration of the l2 solution? 3. If you titrate 100 mL of a drink sample using 3.22 mL of the l2 solution in question 2 above, what is the mg AA/100 mL in the juice? If the label of the pure natural juice is 120% of the RDI per 8 oz. serving, is it in compliance with the federal law?

Explanation / Answer

1.

Serving size = 8 oz = 8 * 29.6 mL

                              = 236.8 mL

120 % Vitamin C i.e. 120 g per 100 mL serving

So mg AA/ 100 mL = 120 * 10^3 = 1.2 * 10^5 mg/ 100 mL serving

2.

Moles of Ascorbic acid = 0.0532 g / 176.12 g/mol

                                     = 3.02 * 10^-4 moles

So moles of Iodine reacted would be half the moles of acid.

Moles of I2 = 1.51 * 10^-4 moles

Concentration of I2 solution = moles /volume (L)

                                              = 1.51 * 10^-4 moles / 0.00522 L

                                             = 0.0289 M

3.

we have calculated molarity of I2 solution as 0.0289 M

Moles of I2 = 0.0289 M * 0.00322 L

                  = 9.31 * 10^-5 moles

o moles of ascorbic acid = 2* 9.31 * 10^-5 = 1.86 * 10^-4 moles

mass of ascorbic acid = 1.86 * 10^-4 moles * 176.12 g/mol

= 32.75 mg

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