Turn in these caleulations with your lab report showing all your work (with unit

Question

Turn in these caleulations with your lab report showing all your work (with units): 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]. (3 pts) 2. Describe how you would prepare 100 ml of 0.08 M sulfuric acid [H2SO4] from an 18 M concentrated sulfuric acid stock solution. (3 pts) 3 Describe how you would prepare 50 ml of a 3% (w/v) soluble starch solution in water. 3 points) Molecular weights of some compounds are presented below: Malonic acid - CHz(COOH)2 104.06 g/mol Manganese sulfate- MnSO4.H2O 169.01 g/mol Hydrogen Peroxide -H202 34.01 g/mol

Explanation / Answer

1. The answer is

Molarity = weight/ Gram molecular weight X Volume in liters.
So, weight = Molarity x Gram molecular weight x volume in liters.

For 50 ml solution =
The required amount of 0.15 M malonic acid =
0.15 X 104.06 X 0.05 = 0.78045 grams.

The required amount of 0.2 M Manganese sulfate =
0.02 X 169.01 X 0.05 = 0.16901 grams.

Dissolove 0.16901 grams of MnSO4 in 20 ml of DDH2O and dissolove 0.78045 grams of malonic acid and adjust the final volume to 50 ml.

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