CHEM PROB MR BOB
posted by Kelli .
I figured all the parts of the problem except for the last one. Can you tell me what to do?
In the preparation of a mustard solution, an individual package of mustard was emptied into a beaker and the mass was determined to be 3.809 grams. The mustard was then dissolved in 50.0 ml. of distilled water. The total mass of the solution was found to be 49.401 grams. Calculate the weight percent of the mustard in the prepared solution.
7.7
The result of a microscale titration showed that 3.91E5 moles of NaOH were required to titrate a 0.775 gram sample of mustard solution to an orangered end point.
Calculate the number of grams of acetic acid in the sample of mustard solution.
2.35×103
Calculate the weight percent of acetic acid in this titrated sample.
0.30
Using the information provided in the questions above, calculate the weight percent of acetic acid in the original package of mustard prepared above???

Is the 7.7 right? All those assumptions I made in my answer yesterday still bother me. However, assuming you have checked a database and things are in order, then 2.35E3 is right.
The second number of 0.30 I would make 0.303 since you're allowed another place.
For the last one, you have 2.35E3 g acetic acid in the titrated sample and the titrated sample contained 0.05976 g (you can round at the end). So
(0.00235/0.05976)*100 = ?% acetic acid in the original sample.
How did I get the 0.05976?
The original was 3.809g/49.401 or 0.07710 g mustard/g soln. You took a 0.775 g sample of the solution which is 0.775 x 0.0771 = 0.05976g
Note: I still don't like the idea that we took a 3.809 g mustard solution, added 50 mL H2O and the TOTAL mass was 49.401 g. It can't be. I might believe a total mass of about 53.2 or so but not 49.401 for a total.
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