A student crushed a vitamin C tablet that weighed 1.532g, and transferred 1.489

Question

A student crushed a vitamin C tablet that weighed 1.532g, and transferred 1.489 g of the powder into a 100mL volumetric flask. After diluting to the 100.00 mL mark, they tot rated a 20.00 mL aliquot of this solution against .0105 M KIO3 (the KIO3) with an excess of HCl and KI. The initial burst reading was .03 mL; the final buret reading was 18.91 mL. Perform the following calculations: 1. Volume of KIO3 used 2. Mmol of KIO3 used 3. mmol of I2 generated 4. Mmol of as orbit acid in the aliquot 5. Mg of as orbit acid in the aliquot 6. Mg of ascot I. Acid in the 100 mL solution 7. Mg of asorbic acid in the original tablet A student crushed a vitamin C tablet that weighed 1.532g, and transferred 1.489 g of the powder into a 100mL volumetric flask. After diluting to the 100.00 mL mark, they tot rated a 20.00 mL aliquot of this solution against .0105 M KIO3 (the KIO3) with an excess of HCl and KI. The initial burst reading was .03 mL; the final buret reading was 18.91 mL. Perform the following calculations: 1. Volume of KIO3 used 2. Mmol of KIO3 used 3. mmol of I2 generated 4. Mmol of as orbit acid in the aliquot 5. Mg of as orbit acid in the aliquot 6. Mg of ascot I. Acid in the 100 mL solution 7. Mg of asorbic acid in the original tablet 1. Volume of KIO3 used 2. Mmol of KIO3 used 3. mmol of I2 generated 4. Mmol of as orbit acid in the aliquot 5. Mg of as orbit acid in the aliquot 6. Mg of ascot I. Acid in the 100 mL solution 7. Mg of asorbic acid in the original tablet

Explanation / Answer

1. volume of KIO3 used = 18.91-0.03 = 18.88 mL

2. mMol of KIO3 = 18.88 * 0.105 = 1.9824 m Mol

3. 1 mole of KIO3 produces 1 mole of KIO, hence

1.9824 m Mol of KIO is produced.

4. m Mol of ascorbic acid = (1.489 / 176.2) m Mol

                                             = 8.45 m Mol in original solution

   8.45 * 100 = x * 20

   x = 0.169 mmol

mass of ascorbic acid = 0.169 *176.12 = 29.76 mg

mass of ascorbic acid in 100 mL of solution = 8.45 * 176.12 = 1488.214 mg

5.

                                             =

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