A student followed the procedure given in the lab exercise to determine the ascorbic acid content in a commercial vitamin C tablet. The students prepared the solution using a 0.460 g vitamin C tablet. The titration required 40.00 mL of the 0.005 M KIO3/KI solution to reach an iodine starch end point. Answer the questions below, using this data.

Calculate the molarity of the iodine solution in millimoles/mL.

To calculate the molarity of the iodine solution in millimoles/mL, we need to use the formula:

Molarity (M) = moles of solute / volume of solution in liters

First, we need to determine the number of moles of iodine (I2) in the titration.

Using the balanced chemical equation:
C6H8O6 (ascorbic acid) + I2 + 2KI → C6H6O6 (dehydroascorbic acid) + 2KCl

We can see that one mole of ascorbic acid reacts with one mole of iodine. Therefore, the number of moles of iodine (I2) in the titration is equal to the number of moles of ascorbic acid in the tablet.

Given that the tablet weighs 0.460 g and we know the molar mass of ascorbic acid is 176.12 g/mol, we can calculate the moles of ascorbic acid:

Moles of ascorbic acid = Mass of tablet / Molar mass of ascorbic acid
= 0.460 g / 176.12 g/mol
= 0.002612 mol

Since the molar ratio between ascorbic acid and iodine is 1:1, the moles of iodine is also 0.002612 mol.

Next, we need to convert the volume of solution used in the titration from milliliters to liters:

Volume of solution (L) = 40.00 mL / 1000 mL/L
= 0.040 L

Finally, we can calculate the molarity of the iodine solution:

Molarity (M) = Moles of iodine / Volume of solution (L)
= 0.002612 mol / 0.040 L
= 0.0653 M

Since the question asks for the molarity in millimoles/mL, we can convert the molarity to millimoles and divide it by the volume in milliliters:

Molarity (mmol/mL) = Molarity (M) x 1000 mmol/mol / Volume of solution (mL)
= 0.0653 M x 1000 mmol/mol / 40.00 mL
= 1.63 mmol/mL

Therefore, the molarity of the iodine solution is 1.63 millimoles per milliliter.