Lab Instructions Graph Fill in a-k please :( Weigh out accurately into a beaker
ID: 837308 • Letter: L
Question
Lab Instructions
Graph
Fill in a-k please :(
Weigh out accurately into a beaker a small amount of bleach powder (0.3-0.4 g) and record the weight. Add 1 M sulfuric acid solution until the fizzing ceases (around 3 mL). This is needed to neutralise the base present. Dilute this accurately to 100 mL| in a volumetric flask with distilled water. Make a second dilution by accurately diluting 1 mL of your first solution to 25 mL. Prepare three 10 mL plastic tubes with 2.5 mL 1 M sulfuric acid and 2.5 mL potassium permanganate solution as before. To each tube add 1 mL of your diluted bleach powder solution. Make up to the 10 mL mark with distilled water and mix thoroughly. Measure the absorbance values as before. Using the average absorbance value from your triplicate analyses, determine: The number of moles of hydrogen peroxide in solution that reacted, using the equation from your previous plot. Calculate the concentration of hydrogen peroxide in your two solutions. Calculate the number of moles of hydrogen peroxide in the active oxygen laundry powder. Work out the % by mass of Na2C03-1.5H202 in the laundry powder.Explanation / Answer
(a) (0.082 + 0.08 + 0.076) / 3 = 0.0793
(b) Plug in the value of y (0.0793) to your best fit equation and you get, x (no. of moles of H2O2) = 1.36E-6
(c) Absorbance = e*c*l where e = molar extinction coefficient = 43.6 M-1cm-1 for H2O2, l = path length = 1 cm (aasume), so concentration, c = absorbance / e*l = 0.0793 / 43.6 *1 = 0.00182 M .
This is the [H2O2] for diluted soln.
(d) V1S1 = V2S2 where V and S denote volume and strength resp.and 1 and 2 denote cocentrated and diluted solution resp.
Then 1 ml x [H2O2] = 25 mL x 0.00182 M
or [H2O2] = 0.0455 M
This is the [H2O2] for original concentrated soln.
(e) No. of moles = molarity (M) x volume (L)
So no. of moles of H2O2 in original soln = 0.0455 x 100 = 4.55
(f) No. of moles of H2SO4 = molarity (M) x volume (L) = 1 x 2.5 x 10-3 = 2.5 x 10-3
H2SO4 reacts with H2O2 in 6:5 molar ratio.
So no. of moles of H2O2 in aliquot = (5/6) x 2.5 x 10-3 = 2.083 x 10-3
(g) Molar mass of H2O2 (or MW) = 34.02 g/mol
(h) Mass of H2O2 in bleech = no. of moles of H2O2 x MW = 2.083 x 10-3 x 34.02 = 70.86 x 10-3 g = 70.86 mg
(h) % mass of H2O2 in bleech = (70.86 x 10-3 / 0.365) x 100 = 19.41 %
(j) Molar mass of Na2CO3,1.5H2O2 is 157.0105 g/mol
(k) Mass of Na2CO3,1.5H2O2 = no. of moles x MW = M x 157.0105 g
% mass of Na2CO3,1.5H2O2 in bleech = (M x 157.0105 / 0.365) x 100 % [Note M is not given]
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