. A student followed the procedure of this experiment to determine the percent N

Question

. A student followed the procedure of this experiment to determine the percent NaOC bleaching solution that was found in the basement of an abandoned house. The in a commercial student diluted 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 faded price label on the gallon bottle read $0.29. The density of the bleaching solution was 1.10 s mL of 0.1052M NaS-O, solution A a. Calculate the number of moles of S,O, ion required for the titration. b. Calculate the number of moles of l2 produced in the titration mixture. c. Calculate the number of moles of OCI ion present in the diluted bleaching solution titrated d. Calculate the mass of NaoCI present in the diluted bleaching solution titrated. e. Determine the volume of commercial bleaching solution present in the diluted bleaching solution titrated.

Explanation / Answer

a) For Na2S2O3 solution, volume = 35.46 ml = 0.03546 L

concentration = 0.1052 M

Moles = concentration*volume = 0.03546*0.1052 = 0.0037 mol

moles of Na2S2O3 = 0.0037 mol

b) Moles of I2 produced:

2S2O32- + I2 ----> 2I- + S4O62-

According to this reaction, 2 moles of S2O32- reacts with 1 mole of I2.

So, moles of I2 = 1/2(moles of S2O32-) = 1/2(0.0037) = 0.00185 mol

c) Moles of OCl- present:

OCl- + 2I- + 2H+ -----> I2 + Cl- + H2O

From this reaction, 1 mole of I2 is prodcued from 1 mole of OCl-.

So, moles of OCl- present = moles of I2 = 0.00185 mol

e) Volume of commercial bleaching powder added = 50.00 ml

Total volume = 250 ml

volume of commercial bleaching powder in diluted solution = 50.00 ml

f) Mass of commercial bleaching powder titrated = density*volume = 1.10*50.00 = 55.00 g

g) Percent NaOCl in commercial bleaching powder:

Moles of OCl- = moles of NaOCl = 0.00185 mol

Molar mass of NaOCl = 74.5 g/mol

Mass of NaOCl = moles*molar mass = 0.00185*74.5 = 0.138 g

% NaOCl = (mass of NaOCl/mass of bleaching powder)*100

= (0.138/55.00)*100 = 0.25 %

h) 1 gallon = 3.785 L

density = 1.10 g/ml = 1.10 kg/L

Mass of 1 gallon = density*volume = 1.10*3.785 = 4.164 kg = 4164 g

i) Price of 1 gallon = price of 4164 g = $ 0.79

Price of 100 g = (0.79/4164)*100 = $0.02

j) 50.00 ml of commercial bleaching powder contains 0.138 g of NaOCl.

100 g of NaOCl will be present in = (50.00/0.138)*100 = 36231.00 ml = 36.231 L

volume = 36.231 L = 36.231/3.785 gallon = 9.57 gallon

1 gallon = $0.79

9.57 gallon = $0.79*9.57 = $7.56

It will cost $7.56

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