Answered: It takes 50.0 ml of 0.0500 M NazS203…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Transcribed Image Text:lews SEC # Evaluation of Commercial Bleaching Solutions 22 DATA SHEET Brand name of commercial bleaching solution: Great Value Container Cost3.60 Container Volume: 2391 Cost per gallon:5.7i 2.5ba Density of commercial bleaching Solution: 1.05 g/ml Trial 1 Trial 2 Trial 3 (if necessary) Molarity of NazS2O3 (stock solution) o,100M 6.10OM Volume of diluted bleach titrated (taken from the volumetric flask) Initial Burette Reading Final Burette Reading Oh' 0५८ Volume of Na2S203 used in the titration (mL) 10.80 Moles of NazS203 used in the titration TI'l J12-13 Moles of S203-2 used 6.83 6.83 Moles of I2 initially present in titration mixture Moles of NaOCI in the diluted bleach titrated G.415 Moles of OC-in the diluted bleach titrated ९४ १ ४४°० 30.99g.30.899 Mass of NaOCI in the diluted bleach titrated Volume of commercial bleach titrated 24.07g 24.079 Mass of NaOCI in commercial bleach titrated Mass of commercial bleaching solution titrated 164.1 429 Voll. 47 % NaOCl in commercial bleaching solution 22.2%% 22.2% Average % NaOCI in commercial bleaching solution TA Initials

It takes 50.0 ml of 0.0500 M NazS203 solution to completely react with the iodine present
in a 125.0 ml iodine solution. How many grams of iodine (I2) were initially present in the
sample?
4)

Expert Solution

Step 1 Given

Molarity of Na₂S₂O₃ solution = 0.0500 M

Volume of Na₂S₂O₃ solution = 50.0 mL

Volume of iodine = 125.0 mL

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