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### 6B.11 Part B Question

Posted: Thu Jan 17, 2019 4:52 pm
It asks for the mass of Na2O added to the first flask, and I looked up the answer in the book as I was confused and I have no idea why the book has the answer:
Question: Added Na2o to a volumetric flask of volume 200 mL, filled w/ water resulting in 200 mL of NaOH solution. 5mL of solution transferred to another volumetric flask and diluted to 500 mL, pH of the diluted solution is 13.25.
- Molar concentration = 18 molxL^-1
B) Mass of Na2O added to the first flask
18 molxL^-1 ( 0.200L) x (1 mole Na2O / 2 mol NaOH) x (61.98 Na2o / 1 mol Na2o) = 110 grams Na2o

### Re: 6B.11 Part B Question

Posted: Thu Jan 17, 2019 5:09 pm
Part B is just stoichiometry; perhaps you're confused because you haven't written out the chemical reaction: Na2O(s) + H2O(l) --> 2NaOH(aq). Using the mole ratios and the initial volume just as you would solve a stoichiometry problem, you would find that there are 1.8 moles of Na2O. If you multiply by the molar mass of Na2O, you would get the answer, 110g.