Question

3. A student followed the procedure of this experiment to determine the percent NaCl in a commercial bleaching solution that was found in the basement of an abandoned house. The student diluted 50.00 mL of the commercial bleaching solution to 250 mL in a volumetric flask, and titrated a 20-ml aliquot of the diluted bleaching solution. The titration required 35.46 mL of 0.1052M Na S,O, solution. A faded price label on the gallon bottle read $0.79. The density of the bleaching solution was 1.10 g ml.

Answer 1

Chemical Reaction Na, S203 3 NaO Ce 3Nace + Naz S20 density mass Volume Sei Commertial bleachin So, mL wcdes to ci fuuted 2 So ML A dd mL thad mean So mL Commercial Soi =Pausent in 250 m2 iluted dilufeat Solfar titation Nous We ake 20 mL than Commertial Sol itrated SoiConteiins 2 o mL SO 2o mL 2 SO density denattyx Volume mass 4 Mass

Reaetoy ACco to chamicas Na S203 3 mole Naoce malc gire that Now mL O1OS2 M Naz S20 3 5.46 uAed we than mole Na S203 Volume in L) x melaity used (35.46x13)x 3.73 X Male

than Mole Nao h teted 20 mL) S 3x (3.73 X13 1.13 X3moie Nao c Suppose PaaRentin Cemmenial Solby ha n masj So mL mass Commercial Sai in 5S Commericl Soi Naoc to diluteo A dol Now wasler and make mL Solution Cdilute 2 5o diluted Soi o.55 Naoc in 2 5o mL mass than 20 mL tak trahion) contalnin =O5S x20 In 2 So

Hcteel) mass Naou (that / +५.५५ MASS Naoc Mola itated Male Naou o44 -4. S.IXlo mele 74.44 than 5.31X1 18-3 3 % by masA

3 7 85 4 gallon m2 duri ty 1o ML nity Mass 3785.41 416 3.3 S Mass 4163.35 4.1633 k2