Experimental Work Safety Note: Wear proper attire, safety glasses and gloves whe

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Experimental Work Safety Note: Wear proper attire, safety glasses and gloves when handling MLchemicals in today's 1. in your notebook, perform the calaulations necessary to determine the amount of 30% (aq in mL that is a 12.5%(ww) needed to prepare solution (Hoa is hydrogen peroxide) 2. The dilution water must be detonized watet or DI water. 3 Prepare this solution ina 400 mL beaker labeled SalutionA 4. The density of 30% Haoa is 1.11 g/cm'. Safety Note: 30%Hydrogen peroxide is corrosive. This strength is NOT the diluted product that you would purchase at a local pharmacy. SolutionB 100 mL of 0.20 M KOhin 0.10 M Hasoe This 100 mL solution of 020M KIO, and 0.10 MH SO. Calculate the number of moles of each reagent required for the 100-mL solution. Kooh is a solid and the sultaricacid isa water solution that is 20 M Hasoe 1. In a 70 mL of DI water and the calculated quantity of KIOb iodate will take several minutes to dissolve. 2. While the KIOais dissolving, determine the volume of 2.0MH sole that must be added to make the final 100 m solution 0.10 MH 3 When the KIO, is nearly dissolved, add the calculated quantity of 20 M HUSOM to the solution. 4 Stir until the KIO sis completelydissolved then remove the str bar Transfer to a graduated cylinder and add DI waterto get 100 m of 6. Transfer back to the 250-mL beaker and label it SolutionB Page 4 Solution C: 100 mL of a solution that is s malonic acid HaChHao), 3.38 mgmL MnsoeHao, and a 300 ppm starch 1. In a 250 mL beaker add approximately 70 mL of DIwater. 2. With stirring add the calculated quantity of malonic acid. 3 Add the caloulated quantity of manganese suitate monohydrate. Stir until both compounds have dissolved. 5. Add the 0.3% stanch into the beaker so that the final 100-mL solution will be 300 ppm starch. Note: (0,3% starch has the equivalent density of water 1 00gkm 6. Add the soktion to a 100 mL graduated cylinder. 7. Add DI water to the 100-mL mark. 8 Transfer back to the 250 mL beaker and label as

Explanation / Answer

Solution C:

Malonic acid = 15.6 mg/mL

So, in 1 mL malonic acid present = 15.6 mg

In 100 mL malonic acid present = 100 x 15.6 mg = 1560 mg = 1.56 g

MnSO4 H2O = 3.38 mg/mL

So, in 1 mL MnSO4 H2O present = 3.38 mg

In 100 mL MnSO4 H2O present = 100 x 3.38 mg = 338 mg

300 ppm starch

Now,

1 ppm = 0.001 mg/mL

300 ppm = 300 x (0.001 mg/mL) = 0.3 mg/mL

So, starch is in 0.3 mg/mL

In 1 mL starch present = 0.3 mg

In 100 mL starch present = 100 x 0.3 mg = 30 mg

Hence in 100 mL of DI water add the following chemicals:

Malonic acid = 1.56 g

MnSO4 H2O = 338 mg

Starch = 30 mg

Solution B:

So, 1 mole of KIO3 in 1000 mL of DI water = 1 M

0.2 mole of KIO3 in 1000 mL of DI water = 0.2 M

(0.2/10) mole of KIO3 in (1000/10) mL of DI water = 0.2 M

0.02 mole of KIO3 in 100 mL of DI water = 0.2 M

Molar mass of KIO3 = 214 g/mol

1 mole of KIO3 = 214 g

0.02 mole of KIO3 = 0.02 x 214 g = 4.28 g

So, mass of KIO3 is 4.28 g in 100 mL of DI water.

For, H2SO4

Now, M1V1 = M2 V2

(0.1 M) x (100 mL) = (2.0 M) x V2

V2 = (0.1 M) x (100 mL) / (2.0 M)

V2 = 5 mL

So, we have to add 5 mL of 2.0 M H2SO4

Molarity = Moles / liter

So, moles of H2SO4 = 2.0 M / 0.005 L = 400 moles

Solution A:

12.5 % solution = (Mass of solute / Mass of solution) x 100

                      = [ (12.5 g of H2O2) / (12.5 g of H2O2 + 87.5 g of H2O2) ] x 100

So, 12.5 %(w/w) means

12.5 g of H2O2 in 100 mL of H2O2 solution.

Now, the stock solution is 30% (w/w)

So,

30 g of H2O2 in 100 mL of solution.

1 g of H2O2 in (100/30) mL of solution.

12.5 g of H2O2 in (12.5) x (100/30) mL of solution.

12.5 g of H2O2 in 41.67 mL of solution.

Volume of 30% (w/w) = 41.67 mL

Density of 30% (w/w) = 1.11 g/ cm3 = 1.11 g/mL

Density = Mass / Volume

Mass = Density x Volume

Mass of 30% (w/w) H2O2 = (1.11 g/mL) x (41.67 mL)

                                           = 46.25 g

So, take 46.25 g of 30% (w/w) H2O2 and add DI water to make it 100 g solution.

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