A solution of I2 was standardized with ascorbic acid. Using a 0.1000-g sample of pure ascorbic acid (C6H8C6), 25.32 ml of I2 were required to reach the starch end point.

What is the molarity of the iodine solution?

Please check to see if I did this right:

(0.1000g Ascorbic Acid/25.32 mlI2)x(1 mole AA/176 g AA)x(1 mole I2/1 mole AA) = 2.244x10^(-5)

The text says that you need one mole of Ascorbic acid to react with 1 mole of I2) I have my doubt with my calculations because I am not sure if you can cancel out grams and ml as I did in the first part.

I will do the problem and let you check it. I'm glad you included the 1:1 ratio for I had forgotten what it was. I assume you meant the formula to be C6H8O6

mols ascorbic acid = grams/molar mass = 0.1000/176.13 = ??

Since it is a 1:1 titration, then mols I2 = the same number of ??

The definition of molarity = mols/L.
You know mols, ?? so divide by 0.02532 L. I find 0.02242 M. Check my work.

I see what I did wrong now, I forgot that Molarity= mols/L, I get the same number for the answer. Thank You!

Well, it seems like you're on the right track, but let me help you clarify a few things. When calculating the molarity of the iodine solution, you need to consider the stoichiometry of the reaction between ascorbic acid and iodine.

The balanced equation for the reaction is:
C6H8O6 + I2 -> 2HI + C6H6O6

From the equation, you can see that 1 mole of ascorbic acid reacts with 1 mole of iodine. Therefore, the moles of iodine can be directly calculated from the moles of ascorbic acid used.

First, let's convert the mass of ascorbic acid to moles:
0.1000 g C6H8O6 x (1 mole C6H8O6 / 176 g C6H8O6) = 0.0005682 moles C6H8O6

Since 1 mole of ascorbic acid reacts with 1 mole of iodine, the moles of iodine present in the solution is also 0.0005682.

Now, let's calculate the molarity of the iodine solution. You used 25.32 ml of the iodine solution, so we need to convert that to liters:
25.32 ml x (1 L / 1000 ml) = 0.02532 L

Finally, we can divide the moles of iodine by the volume in liters to get the molarity:
0.0005682 moles I2 / 0.02532 L = 0.02242 M

So, the molarity of the iodine solution is approximately 0.02242 M.

Remember, chemistry can be a little tricky, but we're here to help you navigate through it!

Your calculations are correct. You can cancel out grams and milliliters in the first part of the calculation because it is a conversion factor.

To find the molarity of the iodine solution, you need to use the equation:

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

In this case, the solute is iodine (I2), and the volume of solution is 25.32 mL or 0.02532 L.
You have already calculated the number of moles of ascorbic acid, so you can use that to determine the number of moles of iodine.

(0.1000 g Ascorbic Acid / 176 g/mol Ascorbic Acid) x (1 mol I2 / 1 mol Ascorbic Acid) = 5.68 x 10^-4 moles I2

Now, you can use the molarity equation to find the molarity (M) of the iodine solution:

M = (5.68 x 10^-4 moles I2) / (0.02532 L) = 0.0224 M

Therefore, the molarity of the iodine solution is 0.0224 M.

To determine the molarity of the iodine (I2) solution, you need to use the stoichiometry of the reaction between ascorbic acid (C6H8C6) and iodine. The balanced equation for this reaction is:

C6H8C6 + I2 → C6H6C6O6 + 2HI

First, let's calculate the number of moles of ascorbic acid (AA) in the 0.1000 g sample:

moles of ascorbic acid = mass of ascorbic acid / molar mass of ascorbic acid
moles of ascorbic acid = 0.1000 g / 176 g/mol (molar mass of ascorbic acid)
moles of ascorbic acid = 0.0005682 mol

Next, using the stoichiometry of the equation, we can calculate the number of moles of iodine (I2) required to react with the given moles of ascorbic acid:

moles of I2 = 1 mole I2 / 1 mole ascorbic acid
moles of I2 = 0.0005682 mol

Now, let's calculate the molarity (M) of the iodine solution using the volume of 25.32 mL (0.02532 L) and the calculated moles of iodine:

molarity of iodine solution = moles of I2 / volume of iodine solution (in liters)
molarity of iodine solution = 0.0005682 mol / 0.02532 L
molarity of iodine solution = 0.0225 M

So, the molarity of the iodine solution is 0.0225 M.

Your calculations are incorrect because you canceled out grams and milliliters in the first part. The canceling out of units is acceptable only when you have the appropriate conversion factors. In this case, moles are needed to calculate the molarity, so the calculations should follow the steps I explained above.