Question

1. A Solution of I₂ was standardized with ascorbic acid (Asc). Using a 0.1032-g sample of pure ascorbic acid, 25.51 mL of I₂ titrant was required to reach the starch point.

a. What is the morality of the iodine solution?

b. What the titer of the iodine solution?

2 . A sample of fresh grapefruit juice was filtered and titrated with the I₂ solution above. A 100-mL sample of the juice took 11.86 mL of the iodine titrant to reach the starch end point.

a. What is the concentration of vitamin C in the juice in mg vitamin C/100 mL of juice?

b. what quantity of this juice will provide the RDI amount of vitamin C ?

No hand Writing!!

Answer 1

1)

Part a

Mass of ascorbic acid = 0.1032 g

Moles = mass/molecular weight

= 0.1032 g / (176.124 g/mol)

= 5.859 x 10^-4 mol

Volume = 25.51 mL x 1L/1000 mL = 0.02551 L

Molarity = moles/Volume

= 5.859 x 10^-4 mol / 0.02551 L

= 0.0229 M

Part b

Titer of solution = Mass of ascorbic acid / volume of I2

= (0.1032 g x 1000mg/g) / (25.51 mL)

= 4.045 mg asc / mL I2

2)

Part a

Titer of solution = (4.045 mg asc / mL I2) x 11.86 mL I2

Concentration of Vitamin C = 47.979 mg

Concentration of Vitamin C / 100 mL juice

= 4797.9 mg/100 mL juice

Part b

recommended daily intake of vitamin C = 2000 mg

Volume of juice = 2000 mg / (4797.9 mg/100 mL juice)

= 41.68 mL