A student has 100.0 mL of an unknown solution of Vitamin C. She removes 1.00 mL of this solution, dilutes it to 50.0 mL, and titrates this second solution with an iodine solution solution that 8.5x1^-4 M. 6.60 mL of iodine solution is required to reach the endpoint.
1. How many moles of iodine were added to the vitamin C solution to reach the endpoint of this titration?
2. If at the end point, the moles of vitamin C and the moles of iodine are equal, how many milligrams of vitamin C were in the original 100.0 mL?
3. What is the molarity of the original unknown vitamin C solution (prior to dilution)?
You would do well to proof your posts because this one is garble.
A. Doesn't the problem say 6.60 mL is required? It's hard to tell from the post.
B. moles vit C = M I2 x L I2 = ?
g vit C titrated = mols vit C x molar mass vit C.
g vit C in original soln = g tirated x 100/1 = ?
C. M original soln = moles/L soln = moles in original soln/0.1L = ?
we are not given the molar mass of the vit. C. how are we supposed to find the milligrams of vit C are in the orginal sample?
Calculate it. Here is a link.
176.12
http://www.google.com/search?q=vitamin+c&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a
to find the total moles of iodine, do you just multiple 8.5x10^-4 by .00660 L?
If that's the M and L, yes. You made a typo there and I can't tell what you used.
to find the molarity of the original vit C do you just divide the moles found in 1 divided by 0.001 L or 1.0 mL?
Nevermind. i figure it out. Thank you so much for your help!
Read my instructions; part C.
C. M original soln = moles/L soln = moles in original soln/0.1L = ?
It's mols/L and you had 100 mL (0.1L) of the original solution.
would 5.61x10^-5 make sense for the molarity for the original unknown sample of vit C solution prior to dilution?
If I didn't make an error, you and I differ by a factor of 100. Did you correct the titrated amount to the original solution by multiplying by 100?
1. mols I2 = M x L = 0.00660 x 8.5E-4 [= 5.61E-6 mols vit C in titrated sample.
2. 5.61E-6 x (100/1) = 5.61E-4 moles in the original solution ( you took a 1 mL aliquot from the 100 mL original).
3. M = moles/L = 5.61E-4/0.1L = 5.61E-3M