1. Describe how you would prepare 50 mL of a solution containing 0.15 M malonic

ID: 787382 • Letter: 1

Question

1. Describe how you would prepare 50 mL of a solution containing 0.15 M malonic acid [CH2(COOH)2] plus 0.02 M manganese sulfate [MnSO4].

2. Describe how you would prepare 100ml of 0.08 M sulfuric acid [H2SO4] from an 18 M concentrated sulfuric acid stock solution.

3. Describe how you would prepare 50 ml of 0.2 M potassium iodate [KIO3] dissolved in the 0.08 M sulfuric acid [H2SO4] that you have prepared above.

4. Describe how you would prepare 25 ml of 3.6 M hydrogen peroxide [h2O2] from a 30% (w/w) stock H2O2 solution (

5. Describe how you would prepare 50 ml of a 3% (w/v) soluble starch solution in water.

Molecular weights:

Malonic acid: 104.6

Manganese sulfate: 169.01

Potassium iodate: 214

Hydrogen peroxide: 34.01

0.15 M malonic acid means in 1 litre, malonic acid present is 0.15mol

In 50mL, malonic acid present = 0.15/1000)*50=7.5*10^-3 mol = 0.7845g

0.02 M manganese sulfate [MnSO4].means in 1 litre, MnSO4 present is 0.02 mol

In 50mL, MnSO4 present = (0.02*50/1000) = 10^-3mol = 0.169g

2) 18 M concentrated sulfuric acid means 18mol sulphuric acid preent in 1 litre

Hence 0.08mol present in 694.4mL

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